#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long powmod(long long a, long long b, long long p) {
  long long res = 1;
  a %= p;
  while (b) {
    if (b & 1) res = res * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void solve() {
  int n, m;
  cin >> n >> m;
  vector<string> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  string ans = v[0];
  for (int j = 0; j < m; j++) {
    char save = ans[j];
    for (char d = 'a'; d <= 'z'; d++) {
      ans[j] = d;
      bool flag = true;
      for (int z = 0; z < n; z++) {
        int cntErrors = 0;
        for (int c = 0; c < m; c++) {
          if (v[z][c] != ans[c]) {
            cntErrors++;
          }
        }
        if (cntErrors > 1) {
          flag = false;
          break;
        }
      }
      if (flag) {
        cout << ans << endl;
        return;
      }
    }
    ans[j] = save;
  }
  cout << "-1" << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please create a solution in CPP to the following problem:

You are given n strings a_1, a_2, …, a_n: all of them have the same length m. The strings consist of lowercase English letters.

Find any string s of length m such that each of the given n strings differs from s in at most one position. Formally, for each given string a_i, there is no more than one position j such that a_i[j] ≠ s[j].

Note that the desired string s may be equal to one of the given strings a_i, or it may differ from all the given strings.

For example, if you have the strings abac and zbab, then the answer to the problem might be the string abab, which differs from the first only by the last character, and from the second only by the first.

Input

The first line contains an integer t (1 ≤ t ≤ 100) — the number of test cases. Then t test cases follow.

Each test case starts with a line containing two positive integers n (1 ≤ n ≤ 10) and m (1 ≤ m ≤ 10) — the number of strings and their length.

Then follow n strings a_i, one per line. Each of them has length m and consists of lowercase English letters.

Output

Print t answers to the test cases. Each answer (if it exists) is a string of length m consisting of lowercase English letters. If there are several answers, print any of them. If the answer does not exist, print "-1" ("minus one", without quotes).

Example

Input


5
2 4
abac
zbab
2 4
aaaa
bbbb
3 3
baa
aaa
aab
2 2
ab
bb
3 1
a
b
c


Output


abab
-1
aaa
ab
z

Note

The first test case was explained in the statement.

In the second test case, the answer does not exist.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
long long powmod(long long a, long long b, long long p) {
  long long res = 1;
  a %= p;
  while (b) {
    if (b & 1) res = res * a % p;
    a = a * a % p;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
void solve() {
  int n, m;
  cin >> n >> m;
  vector<string> v(n);
  for (int i = 0; i < n; i++) {
    cin >> v[i];
  }
  string ans = v[0];
  for (int j = 0; j < m; j++) {
    char save = ans[j];
    for (char d = 'a'; d <= 'z'; d++) {
      ans[j] = d;
      bool flag = true;
      for (int z = 0; z < n; z++) {
        int cntErrors = 0;
        for (int c = 0; c < m; c++) {
          if (v[z][c] != ans[c]) {
            cntErrors++;
          }
        }
        if (cntErrors > 1) {
          flag = false;
          break;
        }
      }
      if (flag) {
        cout << ans << endl;
        return;
      }
    }
    ans[j] = save;
  }
  cout << "-1" << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```