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

import java.util.Scanner;

public class Codeforces1417D {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int t = sc.nextInt();
		
		for(int tt =0 ; tt < t ;tt++) {
			StringBuilder sb = new StringBuilder();
			int n = sc.nextInt() ;
			long a[] = new long[n+ 1];
			long sum = 0;
			for(int i = 1 ; i <= n ; i++) {
				a[i] = sc.nextLong();
				sum += a[i];
			}
			
			if(sum % n != 0) {
				System.out.println(-1);
				continue;
			}
			long count = 0;
			for(int i =2 ; i <= n ; i++) {
				 long remainder = a[i] % i;
				 long toAdd = i - remainder == i?0:i- remainder;
				 if(toAdd != 0) {
					 String s = "1 "+ i + " " + toAdd;
				 sb.append(s+ "\n");
				 count++;
				 }
				 String y = i+ " 1 " + ((a[i] + toAdd )/ i);
				 sb.append(y+"\n");
				 count++;
			}
			
			long ans = sum/n;
			for(int i = 2 ; i <= n ; i++) {
				String s = "1 " + (i ) + " "+ans;
				count++;
				sb.append(s+"\n");
			}
			System.out.println(count);
			System.out.println(sb);
		
		}
		
		sc.close();
	}
	
}