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.*;
import java.util.*;
 
public class sing {
   public static void main(String[] hi) {
      FastIO io = new FastIO();
      int t = io.nextInt();
      while (t --> 0) {
         int n = io.nextInt();
         int[] array = new int[n];
         long sum = 0;
         for (int i=0; i<n; i++) {
            array[i] = io.nextInt();
            sum += array[i];
         }
         long permutation = (n * (n+1))/2;
         if (sum % permutation != 0) {
            io.println("NO");
         }
         else {
            long temp = Long.valueOf(sum/permutation);
            boolean solved = true;
            ArrayList<Long> answer = new ArrayList<Long>();
            long x = array[n-1] - array[0] + temp;
            if (x%n != 0) {
               solved = false;
            }
            x /= n;
            if (x <= 0) {
               solved = false;
            }
            answer.add(x);
            for (int i=1; i<n; i++) {
               int curr = array[i];
               int prev = array[(i+n-2)%(n-1)];
               long y = prev - curr + temp;
               if (y % n != 0) {
                  solved = false;
               }
               y /= n;
               if (y <= 0) {
                  solved = false;
               }
               answer.add(y);
            }
            if (solved) {
               io.println("YES"); 
               for (long i : answer) {
                  io.print(i + " ");
               }
               io.println();
            }
            else {
               io.println("NO");
            }
         }
      }
      io.close();
   }
   static class FastIO extends PrintWriter {
      BufferedReader br;
      StringTokenizer st;
      public FastIO() {
         super(System.out);
         br = new BufferedReader(new InputStreamReader(System.in));
      }
      public String next() {
         try {
            while (st == null || !st.hasMoreTokens()) {
               st = new StringTokenizer(br.readLine());
            }
            return st.nextToken();
         } 
         catch (Exception e) {
            return null;
         }
      }
      public int nextInt() {
         return Integer.parseInt(next());
      }
      public double nextDouble() {
         return Double.parseDouble(next());
      }
      public long nextLong() {
         return Long.parseLong(next());
      }
      public void close() {
         try {
            br.close();
         } 
         catch (IOException e) {
            e.printStackTrace();
         }
         super.close();
      }
   }
}