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#include <bits/stdc++.h>
using namespace std;
#define st first
#define nd second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define BOOST ios_base::sync_with_stdio(0), cin.tie(0)
template<typename T>
using vc = vector<T>;
using ll = long long;
using ld = long double;
using ii = pair<int, int>;

template<int MOD>
struct mod{
	int val;
    mod(ll x = 0){x%=MOD; if(x<0)x+=MOD; val=x;}
    mod operator+(const mod& x) const{return mod(val+x.val);}
    mod operator-(const mod& x) const{return mod(val-x.val);}
    mod operator*(const mod& x) const{return mod(1ll*val*x.val);}
    mod operator/(const mod& x) const{return *this*x.power(MOD-2);}
    mod& operator+=(const mod& x){return *this=*this+x;}
    mod& operator-=(const mod& x){return *this=*this-x;}
    mod& operator*=(const mod& x){return *this=*this*x;}
    mod& operator/=(const mod& x){return *this=*this/x;}
    bool operator==(const mod& x) const{return val==x.val;}
    bool operator!=(const mod& x) const{return val!=x.val;}
    friend std::ostream& operator<<(std::ostream& out, const mod& x){return out << x.val;}
	mod power(int p) const{
		mod x = *this;
		if(p < 0) return x.power(-p).power(MOD-2);
		mod res = {1};
		for(; p; p>>=1){
			if(p&1) res *= x;
			x *= x;
		}
		return res;
	}
};
using mint = mod<(int)1e9+7>;

const int N = 5e6 + 5;
mint f[N];
mint binom(int a, int b){
	if(a < b) return 0;
    return f[a] / (f[b] * f[a-b]);
}
mint pow2[N];

void solve(){
	int n;
	cin >> n;
	vc<int> t(4*n+1);
	bool zero = 0, one = 0, two = 0;
	for(int i=1; i<=2*n; i++){
		cin >> t[i];
		if(t[i] == 0) zero = 1;
		if(t[i] == 1) one = 1;
		if(t[i] == 2) two = 1;
	}
	if(zero && two){
		cout << 0 << "\n";
		return;
	}
	if(zero){
		int swp = t[1];
		for(int i=1; i<2*n; i++) t[i] = t[i+1];
		t[2*n] = swp;
		for(int i=1; i<=2*n; i++) if(t[i] == 0) t[i] = 2;
	}
	for(int i=1; i<=2*n; i++) t[2*n+i] = t[i];
	
	zero = 0, one = 0, two = 0;
	for(int i=1; i<=4*n; i++){
		if(t[i] == 0) zero = 1;
		if(t[i] == 1) one = 1;
		if(t[i] == 2) two = 1;
	}
	assert(!zero);

	mint ans = 0;

	if(one && !two){
		if(n > 1) ans += (mint)n * n * (n-1) * f[4*n-3] * pow2[2*n-3];
		ans += (mint)n * n * f[4*n-3] * pow2[2*n-2];
		cout << ans * 2 << "\n";
		return;
	}

	if(!one && two){
		if(n > 1) ans += (mint)n * (n-1) * f[4*n-2] * pow2[2*n-2];
		ans += (mint)n * f[4*n-2] * pow2[2*n-1];
		cout << ans << "\n";
		return;
	}
	
	vc<int> sc;
	int bad = 0;
	int last = 0;

	auto upd = [&](int i){
		int nxt = i + 2*n;
		if(t[nxt-1] != t[nxt]){
			if((sc.back()&1) == ((nxt-1)&1)){
				bad++;
			}
			sc.pb(nxt-1);
		}
		if(sc[last] <= i){
			if((sc[last]&1) == (sc[last+1]&1)) bad--;
			last++;
		}
	};

	auto skip = [&](int i) -> bool {
		if(bad) return true;
		if((sc.back()&1) == 0) return true;
		if(t[i + 2*n-1] == 2) return true;
		return false;
	};

	for(int i=1; i<2*n; i++){
		if(t[i] != t[i+1]) sc.pb(i);
	}
	for(int i=1; i<(int)sc.size(); i++){
		if((sc[i-1]&1) == (sc[i]&1)) bad++;
	}
	for(int i=1; i<=2*n; i+=2){
		int s = sc.size();
		assert(last != s);
		int fnt = sc[last];

		if(skip(i)){upd(i); upd(i+1); continue;}

		int seq = s - last + 1;
		int afree = 2*n-seq;
		int myseq = seq/2;
		int myfree = n - myseq;
		int myleft = ((fnt-i+1)/2);
		int myright = myfree - myleft;
		int cards = 4*n-seq;

		mint p1 = 0, p2 = 0, p3 = 0, p4 = 0, ways = 0;
		if(myfree >= 1) p1 = (mint)myleft * (myseq+1) * f[cards-2] * pow2[afree-1]; // 4n-3 w zajete
		if(myfree >= 2) p2 = (mint)myleft * (myfree-1) * f[cards-2] * pow2[afree-2]; // 4n-3 w wolne
		if(myright) p3 = (mint)myright * f[cards-1] * pow2[afree-1]; // 4n-2 w wolne na prawo
		if(myseq) p4 = (mint)myseq * f[cards-1] * pow2[afree]; // 4n-2 w zajete na prawo
		ways = p1 + p2 + p3 + p4;
		ans += ways;

		upd(i); upd(i+1);
	}

	cout << ans << "\n";
}


int main(){
	BOOST;

	f[0] = 1;
	for(int i=1; i<N; i++) f[i] = f[i-1] * i;
	pow2[0] = 1;
	pow2[1] = (mint)1/2;
	for(int i=2; i<N; i++) pow2[i] = pow2[i-1] * pow2[1];

	int q;
	cin >> q;
	while(q--) solve();
}