#include <bits/stdc++.h>
#pragma GCC optimize(2)
using namespace std;
const int N = 2e5 + 10;
int T, n, a[N], vis[N], d[N], mi, mx;
signed main() {
  cin >> T;
  while (T--) {
    cin >> n;
    mi = 0x3f3f3f3f, mx = -999;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), vis[a[i]] = i;
    for (int i = 1; i <= n; i++) {
      mi = min(mi, vis[i]), mx = max(mx, vis[i]);
      if (mx - mi == i - 1)
        putchar('1');
      else
        putchar('0');
    }
    puts("");
  }
  return 0;
}
