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 A {
    //--------------------------INPUT READER---------------------------------//
    static class fs {
        public BufferedReader br;
        StringTokenizer st = new StringTokenizer("");

        public fs() { this(System.in); }
        public fs(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }
        String next() {
            while (!st.hasMoreTokens()) {
                try { st = new StringTokenizer(br.readLine()); }
                catch (IOException e) { e.printStackTrace(); }
            }
            return st.nextToken();
        }

        int ni() { return Integer.parseInt(next()); }
        long nl() { return Long.parseLong(next()); }
        double nd() { return Double.parseDouble(next()); }
        String ns() { return next(); }

        int[] na(long nn) {
            int n = (int) nn;
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = ni();
            return a;
        }

        long[] nal(long nn) {
            int n = (int) nn;
            long[] l = new long[n];
            for(int i = 0; i < n; i++) l[i] = nl();
            return l;
        }
    }
    //-----------------------------------------------------------------------//

    //---------------------------PRINTER-------------------------------------//
    static class Printer {
        static PrintWriter w;
        public Printer() {this(System.out);}
        public Printer(OutputStream os) {
            w = new PrintWriter(os);
        }
        public void p(int i) {w.println(i);}
        public void p(long l) {w.println(l);}
        public void p(double d) {w.println(d);}
        public void p(String s) { w.println(s);}
        public void pr(int i) {w.print(i);}
        public void pr(long l) {w.print(l);}
        public void pr(double d) {w.print(d);}
        public void pr(String s) { w.print(s);}
        public void pl() {w.println();}
        public void close() {w.close();}
    }
    //-----------------------------------------------------------------------//

    //--------------------------VARIABLES------------------------------------//
    static fs sc = new fs();
    static OutputStream outputStream = System.out;
    static Printer w = new Printer(outputStream);
    static long lma = Long.MAX_VALUE, lmi = Long.MIN_VALUE;
    static long ima = Integer.MAX_VALUE, imi = Integer.MIN_VALUE;
    static long mod = 1000000007;
    //-----------------------------------------------------------------------//

    //--------------------------ADMIN_MODE-----------------------------------//
    private static void ADMIN_MODE() throws IOException {
        if (System.getProperty("ONLINE_JUDGE") == null) {
            w = new Printer(new FileOutputStream("output.txt"));
            sc = new fs(new FileInputStream("input.txt"));
        }
    }
    //-----------------------------------------------------------------------//

    //----------------------------START--------------------------------------//
    public static void main(String[] args)
            throws IOException {

        ADMIN_MODE();

        int t = sc.ni();while(t-->0)
            solve();


        w.close();
    }

    static void solve() throws IOException {
        int n = sc.ni();
        long[] arr = sc.nal(n);
        long[] res = new long[n];
        long contSum = (((long)n)*(n+1))/2;
        long sum = 0;
        for(long i: arr) {
            sum += i;
        }
        long q = sum/contSum;
        if(sum%contSum!=0) {
            w.p("NO");
            return;
        }

        for(int i = 0; i < n; i++) {
            long minus = arr[(i+1)%n]-q;
            if((arr[i]-minus)%n!=0) {
                w.p("NO");
                return;
            }
            res[n-i-1]=(arr[i]-minus)/n;
            if(res[n-i-1]==0) {
                w.p("NO");
                return;
            }
        }

        w.p("YES");
        for(long i: res) {
            w.pr(i+" ");
        }
        w.pl();
    }

}