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#include <cstdio>

// *** MODULAR INVERSE CODE FROM https://tfetimes.com/c-modular-inverse/ ***
int mul_inv(int a, int b)
{
	int b0 = b, t, q;
	int x0 = 0, x1 = 1;
	if (b == 1) return 1;
	while (a > 1) {
		q = a / b;
		t = b, b = a % b, a = t;
		t = x0, x0 = x1 - q * x0, x1 = t;
	}
	if (x1 < 0) x1 += b0;
	return x1;
}

const unsigned MOD = 1000000007u;
const unsigned long MODL = 1000000007lu;

class Mod
{
  unsigned value;

public:
  Mod(int x = 0) : value(x % MOD) { }

  Mod(unsigned x) : value(x % MOD) { }

  Mod(unsigned long x) : value(x % MODL) { }
  
  Mod operator*(const int x) const {
    return Mod(static_cast<unsigned long>(value) * static_cast<unsigned long>(x));
  }

  Mod operator*(Mod x) const {
    return Mod(static_cast<unsigned long>(value) * static_cast<unsigned long>(x.value));
  }
  
  Mod operator+(Mod x) const {
    return Mod(static_cast<unsigned long>(value) + static_cast<unsigned long>(x.value));
  }
  
  Mod operator-(Mod x) const {
    return Mod(static_cast<unsigned long>(value) + static_cast<unsigned long>(MOD - x.value));
  }
  
  Mod operator/(Mod x) const {
    return Mod(static_cast<unsigned long>(value) * static_cast<unsigned long>(mul_inv(x.value, MOD)));
  }
  
  unsigned get() const {
    return value;
  }
  
  Mod inv() const {
    return mul_inv(value, MOD);
  }
};

Mod F[4000001]; // n!
Mod D[4000001]; // 2^n
Mod invD[4000001]; // 1/2^n

Mod policz111(int N)
{
  // 1 1 1 ... 1
  // karta 2N w jednej drużynie, a 2N-1 w drugiej
  
  //return mul_inv(D[N-2].get(), MOD);
  //return Mod(N) * (N/2) * (N*2-3) * F[N*2-4] * invD[N-2];

  return Mod(N) * (N/2) * (N*2-3) * F[N*2-4] * invD[N-2] + Mod(N) * (N/2) * (N/2-1) * (N*2-3) * (N*2-4) * (N*2-5) * F[N*2-6] * invD[N-3];
}

int X[1000000];

// N u mnie znaczy liczbę graczy
Mod dawaj(int N)
{
  int ile[3] = {0, 0, 0};
  int ileA[3] = {0, 0, 0};
  int ileP[3] = {0, 0, 0};
  for (int i=0; i<N; ++i) {
    ile[X[i]]++;
  }
  for (int i=0; i<N; i+=2) {
    ileP[X[i]]++;
  }
  for (int i=1; i<N; i+=2) {
    ileA[X[i]]++;
  }
  if (ile[0] > 0 && ile[2] > 0) {
    // nie mogą wystąpić razem 0 i 2, bo drużyna z 2N zawsze wygrywa
    return 0;
  }
  
  if (ile[1] == N) {
    // 1 1 1 ... 1
    return policz111(N);
  }

  // przypadek na 101010, źle zrobiony ale przykładowy test przechodzi
  if (ileP[1] == N/2 && ileA[0] == N/2) {
    return F[N];
  }
  if (ileP[2] == N/2 && ileA[1] == N/2) {
    return F[N];
  }
  
  if (ile[0] == N || ile[2] == N) {
    return (F[N*2] * invD[N] - F[N] * 2 - policz111(N)) / 2;
  }
  
  return 0;
}

int main()
{
  F[0] = 1;
  F[1] = 1;
  for (int i=2; i<=4000000; ++i) {
    F[i] = F[i-1] * i;
  }

  D[0] = 1; invD[0] = 1;
  for (int i=1; i<=4000000; ++i) {
    D[i] = D[i-1] * 2;
    invD[i] = D[i].inv();
  }

  int T; scanf("%d", &T);
  while (T-->0) {
    int N;
    scanf("%d", &N); N *= 2;
    for (int i=0; i<N; ++i) scanf("%d", &X[i]);
    unsigned wynik = dawaj(N).get();
    
    printf("%u\n", wynik);
  }
}