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 Main {
    static BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
    static int mod = (int)1e9+7;
    static boolean[] isprime=new boolean[1];
    static {
        for(int i=2;i<isprime.length;i++) isprime[i]=true;
        for(int i=2;i<isprime.length;i++) {
            if(isprime[i]) {
                for(int k = 2; i*k<isprime.length; k++) {
                    isprime[i*k] = false;
                }
            }
        }
    }
    public static long gcd(long a, long b){
        return b == 0 ? a : gcd(b,a%b);
    }
    public static long lcm(long a, long b){
        return a*b / gcd(a,b);
    }
    public static int getT() throws Exception{
        String ss = bf.readLine();
        if (ss.contains(" ")) ss = ss.substring(0, ss.indexOf(" "));
        return Integer.parseInt(ss);
    }
    public static void getA(int[] a) throws Exception {
        String s[] = bf.readLine().split(" ");
        for(int i = 0;i < a.length;i++){
            a[i] = Integer.parseInt(s[i]);
        }
    }
    public static void getA(long[] a) throws Exception {
        String s[] = bf.readLine().split(" ");
        for(int i = 0;i < a.length;i++){
            a[i] = Long.parseLong(s[i]);
        }
    }
    public static void printA(int[] a) throws Exception {
        for(int i : a){
            bw.write(i + " ");
        }
        println();
    }
    public static void println() throws Exception{
        bw.write("\n");
    }

    public static void main(String[] args) throws Exception {
        int t = getT();
        while (t-- > 0) {
            int n = getT();
            long b[] = new long[n];
            getA(b);
            int a[] = new int[n];
            long sum = Arrays.stream(b).sum();
            long k = n*(n+1)/2;
            if(sum % k != 0){
                bw.write("NO");
                println();
            }else {
                sum /= k;
                boolean f = true;
                for(int i = n-1;i > 0;i--){
                    long d = b[i]-b[i-1];
                    if(sum > d && (sum-d) % n == 0) {
                        a[i] = (int) (sum - d) / n;
                    }else{
                        f = false;
                        break;
                    }
                }
                for(int i = 1;i < n;i++){
                    sum -= a[i];
                }
                a[0] = (int)sum;
                if(a[0] <= 0){
                    f = false;
                }
                if(f){
                    bw.write("YES");
                    println();
                    printA(a);
                }else{
                    bw.write("NO");
                    println();
                }
            }
        }
        bw.flush();
    }

}