#include <bits/stdc++.h>
using namespace std;
int t, n;
long long a[10000];
int main() {
  scanf("%d", &t);
  for (; t > 0; t--) {
    long long sum = 0;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      scanf("%lld", a + i);
      sum += a[i];
    }
    if (sum % n != 0) {
      printf("-1\n");
    } else {
      int cnt = 0;
      vector<string> ans;
      char tmp_str[1001];
      for (int i = 1; i < n; i++) {
        if (a[i] % (i + 1) != 0) {
          sprintf(tmp_str, "%d %d %d\0", 1, i + 1, (i + 1) - a[i] % (i + 1));
          a[i] += (i + 1) - a[i] % (i + 1);
          ans.push_back(tmp_str);
          cnt++;
        }
        sprintf(tmp_str, "%d %d %d\0", i + 1, 1, a[i] / (i + 1));
        ans.push_back(tmp_str);
        a[i] = 0;
        cnt++;
      }
      for (int i = 1; i < n; i++) {
        sprintf(tmp_str, "%d %d %d\0", 1, i + 1, sum / n);
        ans.push_back(tmp_str);
        cnt++;
      }
      printf("%d\n", cnt);
      for (string s : ans) {
        printf("%s\n", s.c_str());
      }
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

You are given an array a consisting of n positive integers, numbered from 1 to n. You can perform the following operation no more than 3n times:

  1. choose three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9); 
  2. assign a_i := a_i - x ⋅ i, a_j := a_j + x ⋅ i. 



After each operation, all elements of the array should be non-negative.

Can you find a sequence of no more than 3n operations after which all elements of the array are equal?

Input

The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.

The first line of each test case contains one integer n (1 ≤ n ≤ 10^4) — the number of elements in the array. The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^5) — the elements of the array.

It is guaranteed that the sum of n over all test cases does not exceed 10^4.

Output

For each test case print the answer to it as follows:

  * if there is no suitable sequence of operations, print -1; 
  * otherwise, print one integer k (0 ≤ k ≤ 3n) — the number of operations in the sequence. Then print k lines, the m-th of which should contain three integers i, j and x (1 ≤ i, j ≤ n; 0 ≤ x ≤ 10^9) for the m-th operation. 



If there are multiple suitable sequences of operations, print any of them. Note that you don't have to minimize k.

Example

Input


3
4
2 16 4 18
6
1 2 3 4 5 6
5
11 19 1 1 3


Output


2
4 1 2
2 3 3
-1
4
1 2 4
2 4 5
2 3 3
4 5 1

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int t, n;
long long a[10000];
int main() {
  scanf("%d", &t);
  for (; t > 0; t--) {
    long long sum = 0;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      scanf("%lld", a + i);
      sum += a[i];
    }
    if (sum % n != 0) {
      printf("-1\n");
    } else {
      int cnt = 0;
      vector<string> ans;
      char tmp_str[1001];
      for (int i = 1; i < n; i++) {
        if (a[i] % (i + 1) != 0) {
          sprintf(tmp_str, "%d %d %d\0", 1, i + 1, (i + 1) - a[i] % (i + 1));
          a[i] += (i + 1) - a[i] % (i + 1);
          ans.push_back(tmp_str);
          cnt++;
        }
        sprintf(tmp_str, "%d %d %d\0", i + 1, 1, a[i] / (i + 1));
        ans.push_back(tmp_str);
        a[i] = 0;
        cnt++;
      }
      for (int i = 1; i < n; i++) {
        sprintf(tmp_str, "%d %d %d\0", 1, i + 1, sum / n);
        ans.push_back(tmp_str);
        cnt++;
      }
      printf("%d\n", cnt);
      for (string s : ans) {
        printf("%s\n", s.c_str());
      }
    }
  }
}

```