#include <bits/stdc++.h>
#define MAXN 100
using namespace std;

int n, m;

vector<int> e[MAXN + 10];
int pre[MAXN + 10], suc[MAXN + 10];
int vis[MAXN + 10];

// 找出 S 的所有后继，如果发现包含 S 的环返回 false，否则返回 true
bool bfs(int S) {
    queue<int> q;
    q.push(S); vis[S] = S;
    while (!q.empty()) {
        int sn = q.front(); q.pop();
        for (int fn : e[sn]) {
            if (fn == S) return false;
            if (vis[fn] == S) continue;
            q.push(fn); vis[fn] = S;
            // fn 是 S 的后继，S 也是 fn 的前继
            pre[fn]++; suc[S]++;
        }
    }
    return true;
}

void solve() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) e[i].clear();
    for (int i = 1; i <= m; i++) {
        int x, y; scanf("%d%d", &x, &y);
        e[x].push_back(y);
    }

    memset(pre, 0, sizeof(int) * (n + 3));
    memset(suc, 0, sizeof(int) * (n + 3));
    memset(vis, 0, sizeof(int) * (n + 3));
    for (int i = 1; i <= n; i++) if (!bfs(i)) {
        // 有环，无解
        for (int j = 1; j <= n; j++) printf("0");
        printf("\n");
        return;
    }

    for (int i = 1; i <= n; i++)
        // 前后继都不超过 floor(n / 2)，可以作为中位数
        if (pre[i] <= n / 2 && suc[i] <= n / 2) printf("1");
        else printf("0");
    printf("\n");
}

int main() {
    int tcase; scanf("%d", &tcase);
    while (tcase--) solve();
    return 0;
}