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. 
#include "bits/extc++.h"
using namespace std; using namespace __gnu_pbds; using ll = long long; using pbds = tree<ll, null_type, less<ll>, rb_tree_tag, tree_order_statistics_node_update>; using vll = vector<ll>; using vvll = vector<vll>;
ll MX = 1e6+1, mod = 1e9+7, inf = 1e18; double eps = 1e-9; void setMod(ll m = 998244353){ mod = m; }
#define endl '\n'
#define test ll cases; cin>>cases; for(ll testCase = 1; testCase <= cases; testCase++)
#define mop(a, op, b) ((a%mod op b%mod) + mod)%mod
/* get(array, n); get(n, array1, array2); get(2D array, n, m); get(...); */ void get(ll* a, ll n){ for(ll i = 0; i < n; i++) cin>>a[i]; } void get(vll& a, ll n){ for(ll i = 0; i < n; i++) cin>>a[i]; } void get(ll n, ll* a, ll* b){ for(ll i = 0; i < n; i++) cin>>a[i]>>b[i]; } void get(ll n, vll& a, vll b){ for(ll i = 0; i < n; i++) cin>>a[i]>>b[i]; } void get(ll** a, ll n, ll m){ for(ll i = 0; i < n; i++) for(ll j = 0; j < m; j++) cin>>a[i][j]; } void get(vvll& a, ll n, ll m){ for(ll i = 0; i < n; i++) for(ll j = 0; j < m; j++) cin>>a[i][j]; } 
template<typename T> ostream& operator<<(ostream& out, const vector<T>& v){ for(auto i = v.begin(); i != v.end(); i++) out<<(*i)<<' '; return out; } template<typename T> ostream& operator<<(ostream& out, const set<T>& s){ out<<'['; for(auto i = s.begin(); i != s.end(); i++) out<<(i!=s.begin()?", ":"")<<(*i); return(out<<']'); } template<typename T1, typename T2> ostream& operator<<(ostream& out, const map<T1, T2>& m){ out<<'{'<<endl; for(auto x : m) out<<"    "<<x.first<<" -> "<<x.second<<endl; return(out<<'}'); } template<typename T> T mpow(T x, ll n){ T ans = 1; while(n){ if(n&1) ans = mop(ans, *, x); n >>= 1, x = mop(x, *, x); } return ans; } ll modinv(ll x){ return mpow(x, mod-2); } vll FACT(1, 1); ll ncr(ll n, ll r){ if(n < r) return 0; while(n >= FACT.size()) FACT.push_back(mop(FACT[FACT.size()-1], *, FACT.size())); return mop(FACT[n],*,mop(modinv(FACT[r]),*,modinv(FACT[n-r]))); } ll fact(ll n){ while(n >= FACT.size()) FACT.push_back(mop(FACT[FACT.size()-1], *, FACT.size())); return FACT[n]; }
void zenit(bool tieIO = 1){ cout<<fixed<<setprecision(9); if(tieIO){ ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL); }} mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
 
int main(){
    zenit();
    test{
            ll n;
            cin>>n;
            ll arr[n];
            get(arr,n);
            if(n==1){cout<<"YES"<<endl;cout<<arr[0]<<endl;continue;}
            ll sum=0;
            for(ll i=0;i<n;i++)sum+=arr[i];
            ll val=(n*(n+1))/2;
            ll ratio=sum/val;
            if(ratio<n || sum%val!=0)
            {
                cout<<"NO"<<endl;continue;
            }
            ll res[n]={0};
            bool flag=true;
            for(ll i=0;i<n && flag;i++)
            {
               ll var=arr[i]-arr[(i+1)%n];
               var+=ratio;
                if(var%n!=0 && var==0)
                {cout<<"NO"<<endl;flag =false;}
                res[(i+1)%n]=var/n;
            }
            if(flag)
            {cout<<"YES"<<endl;
            for(ll i=0;i<n;i++)
            cout<<res[i]<<" ";
            cout<<endl;
            }
    }
return 0;
}