#include <bits/stdc++.h>
using namespace std;
struct operation {
  int i, j, x;
  operation(int i, int j, int x) {
    this->i = i;
    this->j = j;
    this->x = x;
  }
};
int mod(int a, int b) { return (a % b + b) % b; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    bool ok = true;
    vector<operation> operations;
    auto Report = [&](int i, int j, int x) {
      operations.emplace_back(i, j, x);
    };
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int sum = accumulate(a.begin(), a.end(), 0);
    if (sum % n != 0) {
      cout << "-1\n";
      continue;
    }
    int find = sum / n;
    for (int i = 1; i < n; i++) {
      int give = mod(-a[i], i + 1);
      Report(0, i, give);
      Report(i, 0, (a[i] + give) / (i + 1));
    }
    for (int i = 1; i < n; i++) {
      Report(0, i, sum / n);
    }
    cout << operations.size() << '\n';
    for (int i = 0; i < operations.size(); i++) {
      cout << operations[i].i + 1 << ' ' << operations[i].j + 1 << ' '
           << operations[i].x << '\n';
    }
  }
}

### Prompt

In CPP, your task is to solve the following problem:

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;
struct operation {
  int i, j, x;
  operation(int i, int j, int x) {
    this->i = i;
    this->j = j;
    this->x = x;
  }
};
int mod(int a, int b) { return (a % b + b) % b; }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    bool ok = true;
    vector<operation> operations;
    auto Report = [&](int i, int j, int x) {
      operations.emplace_back(i, j, x);
    };
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int sum = accumulate(a.begin(), a.end(), 0);
    if (sum % n != 0) {
      cout << "-1\n";
      continue;
    }
    int find = sum / n;
    for (int i = 1; i < n; i++) {
      int give = mod(-a[i], i + 1);
      Report(0, i, give);
      Report(i, 0, (a[i] + give) / (i + 1));
    }
    for (int i = 1; i < n; i++) {
      Report(0, i, sum / n);
    }
    cout << operations.size() << '\n';
    for (int i = 0; i < operations.size(); i++) {
      cout << operations[i].i + 1 << ' ' << operations[i].j + 1 << ' '
           << operations[i].x << '\n';
    }
  }
}

```