// CERC 2012
// Problem A: Kingdoms
// O(n*2^n) solution
// Author: Adam Polak

#include <cstdio>
using namespace std;

const int N = 20;
int n, d[N][N], sum[N];
bool v[1<<N];

void dfs(int k) {
  v[k] = true;
  for (int i=0; i<n; ++i)
    if ((k&(1<<i)) && sum[i] < 0 && !v[k-(1<<i)]) {
      for (int j=0; j<n; ++j)
        if (k & (1<<j))
          sum[j] -= d[i][j];
      dfs(k-(1<<i));
      for (int j=0; j<n; ++j)
        if (k & (1<<j))
          sum[j] += d[i][j];
    }
}

void One() {
  scanf("%d", &n);
  for (int i=0; i<n; ++i)
    for (int j=0; j<n; ++j)
      scanf("%d", &d[i][j]);

  for (int i=0; i<(1<<n); ++i)
    v[i] = false;

  for (int i=0; i<n; ++i)
    sum[i] = 0;
  for (int i=0; i<n; ++i)
    for (int j=0; j<n; ++j)
      sum[j] += d[i][j];

  dfs((1<<n)-1);

  bool any = false;
  for (int i=0; i<n; ++i)
    if (v[1<<i]) {
      printf("%d ", 1+i);
      any = true;
    }
  if (!any) printf("0 ");
  printf("\n");
}

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