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
#include <bits/stdc++.h>
using namespace std;
int arr[300005];
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    long long int sum = 0;
    for (int i = 1; i <= n; i++) {
      cin >> arr[i];
      sum += arr[i];
    }
    int x = sum / n;
    if (sum % n != 0) {
      cout << "-1\n";
      continue;
    }
    vector<pair<pair<int, int>, int>> v;
    for (int i = 2; i <= n; i++) {
      if (arr[i] % i == 0) {
        v.push_back({{i, 1}, arr[i] / i});
      } else {
        v.push_back({{1, i}, i - arr[i] % i});
        v.push_back({{i, 1}, (arr[i] + i - arr[i] % i) / i});
      }
    }
    for (int i = 2; i <= n; i++) {
      v.push_back({{1, i}, x});
    }
    cout << v.size() << "\n";
    for (int i = 0; i < v.size(); i++) {
      cout << v[i].first.first << " " << v[i].first.second << " " << v[i].second
           << "\n";
    }
  }
}
