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.InputStreamReader;
import java.util.StringTokenizer;
import java.util.*;
import java.io.*;
public class Main {
    // Graph
    // prefix sums
    //inputs
    public static void main(String args[])throws Exception{
        Input sc=new Input();
        precalculates p=new precalculates();
        StringBuilder sb=new StringBuilder();
        int t=sc.readInt();
        for(int f=0;f<t;f++){
            int n=sc.readInt();
            long a[]=sc.readArrayLong();
            long d[]=new long[n];
            long sum=0;
            for(int i=0;i<n;i++){
                sum+=a[i];
            }
            long val=(long)n*(long)(n+1);
            val/=2;
            if(sum%val!=0){
                sb.append("NO\n");
                continue;
            }
            sum/=val;
            boolean b=false;
            long s2=0;
            for(int i=1;i<n;i++){
                long temp=(sum-(a[i]-a[i-1]));
                if(temp%n!=0 || temp<=0){
                    b=true;
                    break;
                }else{
                    d[i]=temp/n;
                    s2+=d[i];
                }
            }
            if(b || s2>=sum){
                sb.append("NO\n");
                continue;
            }
            sb.append("YES\n");
            d[0]=sum-s2;
            for(int i=0;i<n;i++){
                sb.append(d[i]+" ");
            }
            sb.append("\n");
        }
        System.out.print(sb);
    }
    /*
    x 2x    x
    3y y    -2y
    2z 3z   z
    b2-b1=(x+y+z-(b2-b1))/3
     */
}
class Input{
    BufferedReader br;
    StringTokenizer st;
    Input(){
        br=new BufferedReader(new InputStreamReader(System.in));
        st=new StringTokenizer("");
    }
    public int[] readArray() throws Exception{
        st=new StringTokenizer(br.readLine());
        int a[]=new int[st.countTokens()];
        for(int i=0;i<a.length;i++){
            a[i]=Integer.parseInt(st.nextToken());
        }
        return a;
    }
    public long[] readArrayLong() throws Exception{
        st=new StringTokenizer(br.readLine());
        long a[]=new long[st.countTokens()];
        for(int i=0;i<a.length;i++){
            a[i]=Long.parseLong(st.nextToken());
        }
        return a;
    }
    public int readInt() throws Exception{
        st=new StringTokenizer(br.readLine());
        return Integer.parseInt(st.nextToken());
    }
    public long readLong() throws Exception{
        st=new StringTokenizer(br.readLine());
        return Long.parseLong(st.nextToken());
    }
    public String readString() throws Exception{
        return br.readLine();
    }
    public int[][] read2dArray(int n,int m)throws Exception{
        int a[][]=new int[n][m];
        for(int i=0;i<n;i++){
            st=new StringTokenizer(br.readLine());
            for(int j=0;j<m;j++){
                a[i][j]=Integer.parseInt(st.nextToken());
            }
        }
        return a;
    }
}
class precalculates{
    public long gcd(long p, long q) {
        if (q == 0) return p;
        else return gcd(q, p % q);
    }
    public int[] prefixSumOneDimentional(int a[]){
        int n=a.length;
        int dp[]=new int[n];
        for(int i=0;i<n;i++){
            if(i==0)
                dp[i]=a[i];
            else
                dp[i]=dp[i-1]+a[i];
        }
        return dp;
    }
    public int[] postSumOneDimentional(int a[]) {
        int n = a.length;
        int dp[] = new int[n];
        for (int i = n - 1; i >= 0; i--) {
            if (i == n - 1)
                dp[i] = a[i];
            else
                dp[i] = dp[i + 1] + a[i];
        }
        return dp;
    }
    public int[][] prefixSum2d(int a[][]){
        int n=a.length;int m=a[0].length;
        int dp[][]=new int[n+1][m+1];
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++){
                dp[i][j]=a[i-1][j-1]+dp[i-1][j]+dp[i][j-1]-dp[i-1][j-1];
            }
        }
        return dp;
    }
    public long pow(long a,long b){
        long mod=1000000007;
        long ans=0;
        if(b<=0)
            return 1;
        if(b%2==0){
            ans=pow(a,b/2)%mod;
            return ((ans%mod)*(ans%mod))%mod;
        }else{
            return ((a%mod)*(ans%mod))%mod;
        }
    }

}
class GraphInteger{
    HashMap<Integer,vertex> vtces;
    class vertex{
        HashMap<Integer,Integer> children;
        public vertex(){
            children=new HashMap<>();
        }
    }
    public GraphInteger(){
        vtces=new HashMap<>();
    }
    public void addVertex(int a){
        vtces.put(a,new vertex());
    }
    public void addEdge(int a,int b,int cost){
        if(!vtces.containsKey(a)){
            vtces.put(a,new vertex());
        }
        if(!vtces.containsKey(b)){
            vtces.put(b,new vertex());
        }
        vtces.get(a).children.put(b,cost);
//        vtces.get(b).children.put(a,cost);
    }
    public boolean isCyclicDirected(){
        boolean isdone[]=new boolean[vtces.size()+1];
        boolean check[]=new boolean[vtces.size()+1];
        for(int i=1;i<=vtces.size();i++) {
            if (!isdone[i] && isCyclicDirected(i,isdone, check)) {
                return true;
            }
        }
        return false;
    }
    private boolean isCyclicDirected(int i,boolean isdone[],boolean check[]){
        if(check[i])
            return true;
        if(isdone[i])
            return false;
        check[i]=true;
        isdone[i]=true;
        Set<Integer> set=vtces.get(i).children.keySet();
        for(Integer ii:set){
            if(isCyclicDirected(ii,isdone,check))
                return true;
        }
        check[i]=false;
        return false;
    }
}
class union_find {
    int n;
    int[] sz;
    int[] par;

    union_find(int nval) {
        n = nval;
        sz = new int[n + 1];
        par = new int[n + 1];
        for (int i = 0; i <= n; i++) {
            par[i] = i;
            sz[i] = 1;
        }
    }

    int root(int x) {
        if (par[x] == x)
            return x;
        return par[x] = root(par[x]);
    }

    boolean find(int a, int b) {
        return root(a) == root(b);
    }

    int union(int a, int b) {
        int ra = root(a);
        int rb = root(b);
        if (ra == rb)
            return 0;
        if(a==b)
            return 0;
        if (sz[a] > sz[b]) {
            int temp = ra;
            ra = rb;
            rb = temp;
        }
        par[ra] = rb;
        sz[rb] += sz[ra];
        return 1;
    }
}






