#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>

using namespace std;

#define int long long

const int N = 105, mod = 1e9 + 7;
int n, m, a[N], ans, f[N][2][N * N], b[N], tot;

int qpow(int a, int b = mod - 2) {
    int res = 1;
    while (b) {
        if (b & 1) (res *= a) %= mod;
        (a *= a) %= mod, b >>= 1;
    }
    return res;
}

void add(int &x, int y) {
    x += y;
    if (abs(x) > 1) x %= mod;
}

void solve() {
    for (int i = 0; i <= tot; i++) {
        int lim = b[i] * (b[i] - 1) / 2 + b[i];
        for (int j = 0; j <= 1; j++) {
            for (int k = 0; k <= lim; k++) {
                f[i][j][k] = 0;
            }
        }
    }
    f[0][0][0] = 1;
    for (int i = 0; i < tot; i++) {
        int lim = (b[i] - 1) * b[i] / 2 + b[i];
        for (int j = 0; j <= 1; j++) {
            for (int k = 0; k <= lim; k++) {
                if (!f[i][j][k]) continue;
                for (int l = i + 1; l <= tot; l++) {
                    int len = b[l] - b[i] - 1;
                    if (a[b[l]] == 1) {
                        add(f[l][j][k + len * (len - 1) / 2 + len], f[i][j][k]);
                        break;
                    } else {
                        add(f[l][!j][k + len * (len - 1) / 2 + len], f[i][j][k]);
                    }
                }
            }
        }
    }
    ans = 0;
    for (int i = tot; i >= 0; i--) {
        int lim = (b[i] - 1) * b[i] / 2 + b[i];
        for (int j = 0; j <= 1; j++) {
            for (int k = 0; k <= lim; k++) {
                int rst = n - b[i];
                rst = rst * (rst - 1) / 2 + rst;
                if (j & 1) {
                    add(ans, -f[i][j][k] * qpow(k + rst, m) % mod);
                } else {
                    add(ans, f[i][j][k] * qpow(k + rst, m) % mod);
                }
            }
        }
        if (a[b[i]] == 1) break;
    }
    ans = (ans % mod + mod) % mod;
}

signed main() {
    int T; scanf("%lld", &T);
    while (T--) {
        scanf("%lld%lld", &n, &m);
        tot = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
            if (a[i]) b[++tot] = i;
        }
        solve();
        printf("%lld\n", ans);
    }
    return 0;
}