n towns are arranged in a circle sequentially. The towns are numbered from 1 to n in clockwise order. In the i-th town, there lives a singer with a repertoire of a_i minutes for each i ∈ [1, n].

Each singer visited all n towns in clockwise order, starting with the town he lives in, and gave exactly one concert in each town. In addition, in each town, the i-th singer got inspired and came up with a song that lasts a_i minutes. The song was added to his repertoire so that he could perform it in the rest of the cities.

Hence, for the i-th singer, the concert in the i-th town will last a_i minutes, in the (i + 1)-th town the concert will last 2 ⋅ a_i minutes, ..., in the ((i + k) mod n + 1)-th town the duration of the concert will be (k + 2) ⋅ a_i, ..., in the town ((i + n - 2) mod n + 1) — n ⋅ a_i minutes.

You are given an array of b integer numbers, where b_i is the total duration of concerts in the i-th town. Reconstruct any correct sequence of positive integers a or say that it is impossible.

Input

The first line contains one integer t (1 ≤ t ≤ 10^3) — the number of test cases. Then the test cases follow.

Each test case consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 4 ⋅ 10^4) — the number of cities. The second line contains n integers b_1, b_2, ..., b_n (1 ≤ b_i ≤ 10^{9}) — the total duration of concerts in i-th city.

The sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, print the answer as follows:

If there is no suitable sequence a, print NO. Otherwise, on the first line print YES, on the next line print the sequence a_1, a_2, ..., a_n of n integers, where a_i (1 ≤ a_i ≤ 10^{9}) is the initial duration of repertoire of the i-th singer. If there are multiple answers, print any of them.

Example

Input


4
3
12 16 14
1
1
3
1 2 3
6
81 75 75 93 93 87


Output


YES
3 1 3 
YES
1 
NO
YES
5 5 4 1 4 5 

Note

Let's consider the 1-st test case of the example:

  1. the 1-st singer in the 1-st city will give a concert for 3 minutes, in the 2-nd — for 6 minutes, in the 3-rd — for 9 minutes; 
  2. the 2-nd singer in the 1-st city will give a concert for 3 minutes, in the 2-nd — for 1 minute, in the 3-rd - for 2 minutes; 
  3. the 3-rd singer in the 1-st city will give a concert for 6 minutes, in the 2-nd — for 9 minutes, in the 3-rd — for 3 minutes. 
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.StringTokenizer;

public class Problem1618E {


    static class FastReader {
        BufferedReader br;
        StringTokenizer st;
        public FastReader() { br = new BufferedReader(new InputStreamReader(System.in)); }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() { return Integer.parseInt(next()); }

        long nextLong() { return Long.parseLong(next()); }

        double nextDouble() { return Double.parseDouble(next()); }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            }
            catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }


    public static void main(String[] args){
        FastReader fr = new FastReader();
        int tests = fr.nextInt();
        while(tests > 0){

            int n = fr.nextInt();
            long[] r = Arrays.stream(fr.nextLine().split(" ")).mapToLong(Long::parseLong).toArray();

            long sum = Arrays.stream(r).sum();
            long k;
            long seq = (long) n * (n+1) / 2;
            if(sum % seq == 0) {
                k = sum / seq;
            } else {
                System.out.println("NO");
                tests--;
                continue;
            }


            long[] a = new long[n];
            boolean hasSol = true;

            for(int i=1; i<n; i++){
                long temp = r[i-1] - r[i] + k;
                if(temp <= 0 || temp % n != 0) {
                    hasSol = false;
                    break;
                }
                a[i] = temp / n;
            }

            if(hasSol){
                a[0] = r[0];
                for(int i=1; i<n; i++){
                    a[0] -= a[i]*(n+1-i);
                }
                if(a[0] > 0) {
                    System.out.println("YES");
                    for(int i=0; i<n; i++){
                        System.out.print(a[i]+" ");
                    }
                    System.out.println();
                }
                else System.out.println("NO");
            } else {
                System.out.println("NO");
            }

            tests--;
        }
    }
}
