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. 
// Generated by Code Flattener.
// https://plugins.jetbrains.com/plugin/9979-idea-code-flattener

import java.io.Closeable;
import java.io.DataInputStream;
import java.io.IOException;

public class E {
    public static void main(String[] args) throws IOException {
        FastScanner scanner = new FastScanner();
        int t = scanner.nextInt();
        cases:
        while (t-- > 0) {
            int n = scanner.nextInt();

            int[] b = new int[n];
            int sumB = 0;
            for (int i = 0; i < n; i++) {
                b[i] = scanner.nextInt();
                sumB += b[i];
            }
            if ((2 * sumB) % (n * (n + 1)) != 0) {
                System.out.println("NO");
                continue cases;
            }
            int sum = 2 * sumB / n / (n + 1);

            int[] d = new int[n];
            d[0] = b[0] - b[n - 1];
            int max = Integer.MIN_VALUE;
            for (int i = 1; i < n; i++) {
                d[i] = b[i] - b[i - 1];
                max = Math.max(max, d[i]);
            }
            StringBuilder builder = new StringBuilder();
            for (int i = 0; i < n; i++) {
                int val = sum - d[i];
                if (val <= 0 || val % n != 0) {
                    System.out.println("NO");
                    continue cases;
                }
                builder.append(val / n);
                if (i != n - 1) builder.append(" ");
            }
            System.out.println("YES");
            System.out.println(builder);
        }
    }

    private static class FastScanner implements Closeable {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;

        public FastScanner() {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public int nextInt() throws IOException {
            int ret = 0;
            byte c = read();
            while (c <= ' ') c = read();
            boolean neg = (c == '-');
            if (neg) c = read();
            do {
                ret = ret * 10 + c - '0';
            } while ((c = read()) >= '0' && c <= '9');
            if (neg) return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1) buffer[0] = -1;
        }

        private byte read() throws IOException {
            if (bufferPointer == bytesRead) fillBuffer();
            return buffer[bufferPointer++];
        }

        @Override
        public void close() throws IOException {
            if (din == null) return;
            din.close();
        }
    }
}