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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define st first
#define nd second
ll mod = 1000000007;

ll fact[4000005];
ll n;
ll P;
ll powmod(ll x, ll k) {
    ll res = 1;
    for (ll y = x; k; k>>=1, y = (y*y)%mod) {
        if (k&1) res = (res*y)%mod;
    }
    return res;
}

ll inv(ll x) {
    return powmod(x, mod-2);
}

ll only_twos(int n) {
    ll ans = (2*n*(2*n-1))%mod;
    ans = (ans * fact[4*n-2])%mod;
    ans = (ans * P)%mod;
    return ans;
}
ll only_ones(int n) {
    ll ans = (2*n*2*n*(2*n-1))%mod;
    ans = (ans * fact[4*n-3])%mod;
    ans = (ans * P)%mod;
    ans = (ans+ans)%mod;
    return ans;
}

ll type1(vector<int>& segments) {
    int k = segments.size();
    ll total = 0;
    for (int i = 0; i < segments.size(); i+=2) { // ustalamy segment 1, na końcu będzie A
        ll cur = 2*n-((segments[i+1]/2)*2)-(k/2);
        cur = (cur * fact[4*n-k-1])%mod;
        total += cur;
    }
    total %= mod;
    total = (total * P)%mod;
    total = (total*powmod(2, k))%mod;
    return total;
}

ll type2(vector<int>& segments) {
    int k = segments.size();
    ll total = 0;
    for (int i = 0; i < segments.size(); i+=2) { // ustalamy segment 1, na końcu będzie A
        ll cur = (((segments[i+1]/2)*2) * (2*n-(k/2)-1))%mod;
        cur = (cur * fact[4*n-k-2])%mod;
        total += cur;
    }
    total %= mod;
    total = (total * P)%mod;
    total = (total*powmod(2, k))%mod;
    return total;
}

ll type3(vector<int>& segments) {
    int k = segments.size();
    ll total = 0;
    for (int i = 0; i < segments.size(); i+=2) { // ustalamy segment 1, na końcu będzie 10 a w środku A
        for (int j = 0; j < segments[i]-1; j+=2) {
            ll cur = (2*n-(k/2)-((segments[i]-1-j)/2)*2);
            cur = (cur*fact[4*n-k-2])%mod;
            total += cur;
        }
    }
    total %= mod;
    total = (total * P)%mod;
    total = (total*powmod(2, k+1))%mod;
    return total;
}

ll type4(vector<int>& segments) {
    int k = segments.size();
    ll total = 0;
    for (int i = 0; i < segments.size(); i+=2) { // ustalamy segment 1, na końcu będzie 10 a w środku A, a pomiędzy A i 10 będzie 9
        for (int j = 0; j < segments[i]-1; j+=2) {
            ll cur = (((segments[i]-1-j)/2)*2 * (2*n-(k/2)-1))%mod;
            cur = (cur*fact[4*n-k-3])%mod;
            total += cur;
        }
    }
    total %= mod;
    total = (total * P)%mod;
    total = (total*powmod(2, k+1))%mod;

    return total;
}

void solve() {
    cin >> n;
    P = inv(powmod(2, 2*n));
    vector<int> s(2*n);
    bool ex[3] = {false, false, false};
    for (int i = 0; i < 2*n; i++) {
        cin >> s[i];
        // cout << s[i] << '\n';
        ex[s[i]] = true;
        if (s[i] == 0) s[i] = 2;
    }
    if (ex[0] && ex[2]) {
        cout << "0\n";
        return;
    }
    int ace_pos;
    if (ex[0]) ace_pos = 1;
    else ace_pos = 0;
    ex[2] = (ex[0] || ex[2]);
    if (ex[1] == false) {
        cout << only_twos(n) << '\n';
        return;
    }
    if (ex[2] == false) {
        cout << only_ones(n) << '\n';
        return;
    }
    int start = 0;
    while (s[start] == s.back()) {
        s.push_back(s[start]);
        start++;
    }
    while (s[start] == 2) {
        s.push_back(2);
        start++;
    }
    if (start&1) {
        if (ace_pos == 1) {
            cout << "0\n";
            return;
        }
    }
    else {
        if (ace_pos == 0) {
            cout << "0\n";
            return;
        }
    }
    vector<int> segments;
    int len = 1;
    for (int i = start+1; i < s.size(); i++) {
        if (s[i] != s[i-1]) {
            segments.push_back(len);
            len=1;
        }
        else len++;
    }
    segments.push_back(len);
    for (int x: segments) {
        if (!(x&1)) {
            cout << "0\n";
            return;
        }
    }
    ll ans = (type1(segments) + type2(segments) + type3(segments) + type4(segments))%mod;
    cout << ans << '\n';
    return;
}


int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int t = 1;
    fact[0] = 1;
    for (ll i = 1; i <= 4000000; i++) fact[i] = (fact[i-1]*i)%mod;
    cin >> t;
    while (t--) {
        solve();
    }
}