#include <bits/stdc++.h>
template <class T>
inline void read(T &x) {
  char c = getchar(), f = 0;
  x = 0;
  while (!isdigit(c)) f = (c == '-'), c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x = f ? -x : x;
}
using namespace std;
const int N = 10001;
const int M = 100001;
const int HA = 998244353;
int n, a[M];
void solve() {
  for (int i = (1); i <= (n); i++) a[i] = 0;
  read(n);
  long long sum = 0;
  for (int i = (1); i <= (n); i++) {
    read(a[i]);
    sum += a[i];
  }
  if (sum % n) {
    puts("-1");
    return;
  }
  long long ans = (n - 1) * 2;
  for (int i = (2); i <= (n); i++) {
    if (a[i] % i) ans++;
  }
  printf("%lld\n", ans);
  for (int i = (2); i <= (n); i++) {
    if (a[i] % i) {
      long long x = i - a[i] % i;
      printf("%d %d %d\n", 1, i, x);
      a[i] += x, a[1] -= x;
    }
    printf("%d %d %d\n", i, 1, a[i] / i);
    a[1] += a[i], a[i] = 0;
  }
  for (int i = (2); i <= (n); i++) printf("%d %d %d\n", 1, i, a[1] / n);
}
int main() {
  int t;
  read(t);
  while (t--) solve();
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
template <class T>
inline void read(T &x) {
  char c = getchar(), f = 0;
  x = 0;
  while (!isdigit(c)) f = (c == '-'), c = getchar();
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x = f ? -x : x;
}
using namespace std;
const int N = 10001;
const int M = 100001;
const int HA = 998244353;
int n, a[M];
void solve() {
  for (int i = (1); i <= (n); i++) a[i] = 0;
  read(n);
  long long sum = 0;
  for (int i = (1); i <= (n); i++) {
    read(a[i]);
    sum += a[i];
  }
  if (sum % n) {
    puts("-1");
    return;
  }
  long long ans = (n - 1) * 2;
  for (int i = (2); i <= (n); i++) {
    if (a[i] % i) ans++;
  }
  printf("%lld\n", ans);
  for (int i = (2); i <= (n); i++) {
    if (a[i] % i) {
      long long x = i - a[i] % i;
      printf("%d %d %d\n", 1, i, x);
      a[i] += x, a[1] -= x;
    }
    printf("%d %d %d\n", i, 1, a[i] / i);
    a[1] += a[i], a[i] = 0;
  }
  for (int i = (2); i <= (n); i++) printf("%d %d %d\n", 1, i, a[1] / n);
}
int main() {
  int t;
  read(t);
  while (t--) solve();
  return 0;
}

```