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.*;

public class CF1 {
    public static void main(String[] args) {
        FastScanner sc=new FastScanner();
        int T=sc.nextInt();
//        int T=1;
        for (int tt=0; tt<T; tt++){
            int n = sc.nextInt();
            long sum=0L;
            int arr[]= sc.readArray(n);
            for (int i=0; i<n; i++){
                sum+=arr[i];
            }
            boolean possible =true;
            long y=(long)((long)n*(long)(n+1))/2;
            if (sum%y!=0) possible=false;
            sum/=y;

//            if (sum<n)

            int ans[]= new int [n];
            for (int i=0; i<n-1; i++){
                int x=arr[i]+(int)sum-arr[i+1];
                if (x<=0 || x%n!=0) {
                    possible=false;
                    break;
                }
                else ans[i+1]=x/n;
            }
            int x =arr[n-1]+(int)sum-arr[0];
            if (x<=0 || x%n!=0) {
                possible=false;

            }
            else ans[0]=x/n;
            if (possible){
                System.out.println("YES");
                StringBuilder sb= new StringBuilder();
                for (int i=0; i<n; i++){
                    sb.append(ans[i]+" ");
                }
                System.out.println(sb.toString());
            }
            else System.out.println("NO");
        }
// arr[i+1]= arr[0]+sum-n*x;
    }
    static boolean help(char arr[], int start, int end){
        for (int i=start+1; i<end; i++){
            if (arr[i]=='b' && arr[i-1]=='a' && arr[i+1]=='c') return true;
        }
        return false;
    }
    static long mod =998244353L;
    static long power (long a, long b){
        long res=1;
        while (b>0){
            if ((b&1)== 1){
                res= (res * a % mod)%mod;
            }
            a=(a%mod * a%mod)%mod;
            b=b>>1;
        }
        return res;
    }
    boolean[] sieveOfEratosthenes(int n)
    {

        boolean prime[] = new boolean[n+1];
        for(int i=0;i<=n;i++)
            prime[i] = true;

        for(int p = 2; p*p <=n; p++)
        {
            if(prime[p] == true)
            {
                for(int i = p*p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;
    }
    static void sort(int[] a) {
        ArrayList<Integer> l=new ArrayList<>();
        for (int i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }
    static void sortLong(long[] a) {
        ArrayList<Long> l=new ArrayList<>();
        for (long i:a) l.add(i);
        Collections.sort(l);
        for (int i=0; i<a.length; i++) a[i]=l.get(i);
    }
    static long gcd (long n, long m){
        if (m==0) return n;
        else return gcd(m, n%m);
    }
    static class Pair{
        int x,y;
        public Pair(int x, int y){
            this.x = x;
            this.y = y;
        }
    }
    static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}