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#include <bits/stdc++.h>

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define FORD(i, a, b) for(int i = (a); i >= (b); --i)
#define VAR(v, i) __typeof(i) v=(i)
#define FORE(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)
#define all(v) (v).begin(),(v).end()

#define PII pair<int,int>
#define st first
#define nd second
#define pb push_back
#define lint long long int
#define VI vector<int>

#define debug(x) {cout <<#x <<" = " <<x <<endl; }
#define debug2(x,y) {cerr <<#x <<" = " <<x << ", "<<#y<<" = "<< y <<endl; } 
#define debug3(x,y,z) {cerr <<#x <<" = " <<x << ", "<<#y<<" = "<< y << ", " << #z << " = " << z <<endl; } 
#define debugv(x) {{cout <<#x <<" = "; FORE(itt, (x)) cerr <<*itt <<", "; cerr <<endl; }}
#define debugt(t,n) {{cerr <<#t <<" = "; FOR(it,0,(n)) cerr <<t[it] <<", "; cerr <<endl; }}

#define make( x) int (x); scanf("%d",&(x));
#define make2( x, y) int (x), (y); scanf("%d%d",&(x),&(y));
#define make3(x, y, z) int (x), (y), (z); scanf("%d%d%d",&(x),&(y),&(z));
#define make4(x, y, z, t) int (x), (y), (z), (t); scanf("%d%d%d%d",&(x),&(y),&(z),&(t));
#define makev(v,n) VI (v); FOR(i,0,(n)) { make(a); (v).pb(a);} 
#define IOS ios_base::sync_with_stdio(0)
#define HEAP priority_queue

#define read( x) scanf("%d",&(x));
#define read2( x, y) scanf("%d%d",&(x),&(y));
#define read3(x, y, z) scanf("%d%d%d",&(x),&(y),&(z));
#define read4(x, y, z, t) scanf("%d%d%d%d",&(x),&(y),&(z),&(t));
#define readv(v,n) FOR(i,0,(n)) { make(a); (v).pb(a);}
#define jeb() fflush(stdout);

using namespace std;


void normalizuj(vector < int >&tab, int n) {
  int ile_jedynek = 0;
  FOR(i, 0, n + 5) {
    if (tab[i] == 0) {
      if (ile_jedynek == 0) {
        ile_jedynek = 0;
        continue;
      }                         // 0 -> 0
      if (ile_jedynek == 1) {
        ile_jedynek = 0;
        continue;
      }                         // 010 -> 010
      if (ile_jedynek % 2 == 0) { // 01^(2k) -> 00 (01)^k
        FOR(j, 1, ile_jedynek + 1) {
          tab[i - j] = 0;
          if (j % 2 == 0)
            tab[i - j] = 1;
        }
        tab[i - ile_jedynek] = 0;
        tab[i] = 1;
        ile_jedynek = 1;
        continue;
      }
      // 01^(2k+1) -> 010(01)^k
      FOR(j, 1, ile_jedynek+1) {
        tab[i - j] = 0;
        if (j % 2 == 0)
          tab[i - j] = 1;
      }
      tab[i] = 1;
      tab[i - ile_jedynek] = 1;
      tab[i - ile_jedynek + 1] = 0;
      ile_jedynek = 1;
      continue;
    }
    if (tab[i] == 1) {
      ile_jedynek++;
    }
    if (tab[i] == 2) {          // nastepna zawsze jest == 0
      if (ile_jedynek == 0) {
        continue;
      }
      if (ile_jedynek == 1) {
        tab[i - 1] = 0;
        tab[i] = 1;
        tab[i + 1] = 1;
        ile_jedynek = 1;
        // 0120 -> 0011
      }
      if (ile_jedynek == 2) {
        tab[i - 1] = 0;
        tab[i] = 1;
        tab[i + 1] = 1;
        ile_jedynek = 1;
        // 01120 -> 01011
      }
      // innych mozliwosci nie MA !!!
    }
  }
  return;
}

void eliminate_102_002(vector < int >&tab, int n) {
  // chcemy wyeliminowac sekwencje 102 i 002
  // ciag na wejsciu jest znormalizowany.
  // i kazda 2 jest "otoczona" zerami
  FORD(i, n + 5, 2) {
    if (tab[i] == 2) {
      if ((tab[i - 1] == 0) && (tab[i - 2] == 1)) {
        // 102 -> 211
        tab[i - 2] = 2;
        tab[i - 1] = 1;
        tab[i] = 1;
        continue;
      }
      if ((tab[i - 1] == 0) && (tab[i - 2] == 0)) {
        // 002 -> 111
        tab[i - 2] = 1;
        tab[i - 1] = 1;
        tab[i] = 1;
      }
    }
  }
  if ((tab[1] == 2) && (tab[0] == 0)) {
    tab[1] = 1;
    tab[0] = 2;
  }
  // teraz po kazdej dwojce jest pewna (niezerowa) ilosc jedynek
  return;
}

// usuwa 2 z poczatku
void eliminate_front(vector < int >&tab, int n){
  if ((tab[0] == 2) && (tab[1] == 0)) {
    tab[0] = 0;
    tab[1] = 1;
  }

  int i = 0;
  if (tab[i] == 2) {
    int ile_jedynek = 0;
    while (tab[i + ile_jedynek + 1] == 1)
      ile_jedynek++;
    if ((ile_jedynek != 0) && (ile_jedynek % 2 == 0)) {
      // 21^(2k)0 -> (01)^(k+1)
      FOR(j,0, ile_jedynek + 2) {
        if (j % 2 == 0)
          tab[i + j] = 0;
        if (j % 2 == 1)
          tab[i + j] = 1;
      }
      i += ile_jedynek + 1;
    }
    if (ile_jedynek % 2 == 1) {
      // 21^(2k+1)0 -> 1(01)^(k+1)
      FOR(j,0, ile_jedynek + 2) {
        if (j % 2 == 0)
          tab[i + j] = 1;
        if (j % 2 == 1)
          tab[i + j] = 0;
      }
      i += ile_jedynek + 1;
    }
  }
  // poniewaz w danych nie bylo 002 to w wyniku nie bedzie 102
}

void eliminate_last(vector < int >&tab, int n) {
  // ostatnia eliminacja
  FOR(i, 2, n + 6) {
    if (tab[i] == 2) {
      // 1102  -> 0012
      if ((tab[i - 1] == 0) && (tab[i - 2] == 1) && (tab[i - 3] == 1)) {
        tab[i - 3] = 0;
        tab[i - 2] = 0;
        tab[i - 1] = 1;
        tab[i] = 2;
        continue;
      }
    }
    if (tab[i - 1] == 2) {
      if (tab[i] == 0) {
        // 120 -> 011
        tab[i - 2] = 0;
        tab[i - 1] = 1;
        tab[i] = 1;
        continue;
      }
      if (tab[i] == 1) {
        // 121 --> 012
        tab[i - 2] = 0;
        tab[i - 1] = 1;
        tab[i] = 2;
        continue;
      }
    }
  }
}

void dodaj(VI& v, VI& w) {
	int n = max(v.size(), w.size());
	while (v.size() < n + 10) v.push_back(0);
	FOR(i,0,w.size()) v[i] += w[i];
	normalizuj(v, n);
  eliminate_102_002(v, n);
  eliminate_front(v, n);
  eliminate_last(v, n);
  normalizuj(v, n);
}

void rob_wektor(int a, int b, VI& res) {
	if (a == 1) {
		res[b-1] = 1;
		return;
	}
	if (a == 2) {
		if (b == 2) {
			res[0] = 1;
			res[2] = 1;
			return;
		}
		res[b] = 1;
		res[b-3] = 1;
		return;
	}
	rob_wektor(a-2, b-2, res);
	res[a+b-2] = 1;
}

void solve() {
	make(k);
	makev(v, k);
	make(l);
	makev(w, l);
	vector<VI> skladniki;
	FOR(i,0,k) FOR(j,0,l) {
		if (v[i] == 1 && w[j] == 1) {
			int a = min(i, j) + 1;
			int b = max(i, j) + 1;
			VI kij(a+b+5);
			rob_wektor(a, b, kij);
			while (kij.back() == 0) kij.pop_back();
			skladniki.pb(kij);
		}
	}
	VI ans = skladniki[0];
	FOR(i,1,skladniki.size()) {
		dodaj(ans, skladniki[i]); 
	}
	while (ans.back() == 0) ans.pop_back();
	printf("%d ", (int)ans.size());
	FOR(i,0,ans.size()) {
		printf("%d ", ans[i]);
	}
	printf("\n");
}

int main() {
	make(z);
	while (z--) {
		solve();
	}
}