#include <bits/stdc++.h>
using namespace std;
void ha() { cout << "YES" << endl; }
void na() { cout << "NO" << endl; }
long long power(long long a, long long b) {
  long long c, d;
  if (b == 0) {
    return 1 % 1000000007;
  } else if (b == 1) {
    return a % 1000000007;
  } else if (b % 2 == 0) {
    c = power(a, b / 2);
    return (c * c) % 1000000007;
  } else {
    c = power(a, b / 2);
    d = (c * c) % 1000000007;
    d = (d * a) % 1000000007;
    return d;
  }
}
long long plainPower(long long a, long long b) {
  long long sum = 1, i;
  for (i = 1; i <= b; i++) {
    sum *= a;
  }
  return sum;
}
int absolute(int a) {
  if (a >= 0) {
    return a;
  } else {
    return (a * (-1));
  }
}
long long absolute(long long a) {
  if (a >= 0) {
    return a;
  } else {
    return (a * (-1));
  }
}
bool isPrime(int a) {
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      return false;
    }
  }
  return true;
}
bool isTriangle(long long x1, long long y1, long long x2, long long y2,
                long long x3, long long y3) {
  long long area = (x1 * (y2 - y3)) + (x2 * (y3 - y1)) + (x3 * (y1 - y2));
  if (area) {
    return true;
  } else {
    return false;
  }
}
long long findRangeSum(long long hi, long long lo) {
  long long j = (hi * (hi + 1)) / 2;
  long long k = (lo * (lo + 1)) / 2;
  return j - k;
}
bool present(vector<int> &a, int n, int target) {
  int lo = 0, hi = n - 1, mid;
  while (lo <= hi) {
    mid = lo + ((hi - lo) / 2);
    if (a[mid] == target) {
      return true;
    } else if (a[mid] > target) {
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  return false;
}
int Left_bound(vector<int> &a, int n, int target) {
  int lo = 0, hi = n - 1, mid, ans = -1;
  while (lo <= hi) {
    mid = lo + ((hi - lo) / 2);
    if (a[mid] <= target) {
      ans = mid;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return ans + 1;
}
int Right_bound(vector<int> &a, int n, int target) {
  int lo = 0, hi = n - 1, mid, ans = n;
  while (lo <= hi) {
    mid = lo + ((hi - lo) / 2);
    if (a[mid] >= target) {
      ans = mid;
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  return ans + 1;
}
void debugger() {}
int cnt;
int target;
int dfs(vector<vector<int> > &adj, vector<int> &vis, vector<int> &val, int strt,
        int parent) {
  vis[strt] = val[strt];
  int xr = 0;
  for (int i = 0; i < adj[strt].size(); i++) {
    if (adj[strt][i] != parent) {
      xr = (xr ^ dfs(adj, vis, val, adj[strt][i], strt));
    }
  }
  vis[strt] = (xr ^ vis[strt]);
  if (vis[strt] == target) {
    cnt++;
    vis[strt] = 0;
  }
  return vis[strt];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long term = (n * (n + 1)) / 2;
    vector<long long> a(n + 1);
    vector<long long> temp(n + 1);
    long long i, j, k, flag = 0, sum = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
      temp[i] = 0;
    }
    if (sum % term) {
      na();
      continue;
    }
    long long tt = sum / term;
    vector<long long> arr;
    for (i = 1; i < n; i++) {
      arr.push_back(a[i] - a[i - 1]);
    }
    arr.push_back(a[0] - a[n - 1]);
    for (i = 1; i <= n; i++) {
      j = i % n;
      temp[j] = tt - arr[i - 1];
      if (temp[j] % n) {
        flag = 1;
        break;
      }
      temp[j] /= n;
      if (temp[j] <= 0) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      na();
    } else {
      ha();
      for (i = 0; i < n; i++) {
        cout << temp[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Please formulate a Cpp solution to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void ha() { cout << "YES" << endl; }
void na() { cout << "NO" << endl; }
long long power(long long a, long long b) {
  long long c, d;
  if (b == 0) {
    return 1 % 1000000007;
  } else if (b == 1) {
    return a % 1000000007;
  } else if (b % 2 == 0) {
    c = power(a, b / 2);
    return (c * c) % 1000000007;
  } else {
    c = power(a, b / 2);
    d = (c * c) % 1000000007;
    d = (d * a) % 1000000007;
    return d;
  }
}
long long plainPower(long long a, long long b) {
  long long sum = 1, i;
  for (i = 1; i <= b; i++) {
    sum *= a;
  }
  return sum;
}
int absolute(int a) {
  if (a >= 0) {
    return a;
  } else {
    return (a * (-1));
  }
}
long long absolute(long long a) {
  if (a >= 0) {
    return a;
  } else {
    return (a * (-1));
  }
}
bool isPrime(int a) {
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) {
      return false;
    }
  }
  return true;
}
bool isTriangle(long long x1, long long y1, long long x2, long long y2,
                long long x3, long long y3) {
  long long area = (x1 * (y2 - y3)) + (x2 * (y3 - y1)) + (x3 * (y1 - y2));
  if (area) {
    return true;
  } else {
    return false;
  }
}
long long findRangeSum(long long hi, long long lo) {
  long long j = (hi * (hi + 1)) / 2;
  long long k = (lo * (lo + 1)) / 2;
  return j - k;
}
bool present(vector<int> &a, int n, int target) {
  int lo = 0, hi = n - 1, mid;
  while (lo <= hi) {
    mid = lo + ((hi - lo) / 2);
    if (a[mid] == target) {
      return true;
    } else if (a[mid] > target) {
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  return false;
}
int Left_bound(vector<int> &a, int n, int target) {
  int lo = 0, hi = n - 1, mid, ans = -1;
  while (lo <= hi) {
    mid = lo + ((hi - lo) / 2);
    if (a[mid] <= target) {
      ans = mid;
      lo = mid + 1;
    } else {
      hi = mid - 1;
    }
  }
  return ans + 1;
}
int Right_bound(vector<int> &a, int n, int target) {
  int lo = 0, hi = n - 1, mid, ans = n;
  while (lo <= hi) {
    mid = lo + ((hi - lo) / 2);
    if (a[mid] >= target) {
      ans = mid;
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
  return ans + 1;
}
void debugger() {}
int cnt;
int target;
int dfs(vector<vector<int> > &adj, vector<int> &vis, vector<int> &val, int strt,
        int parent) {
  vis[strt] = val[strt];
  int xr = 0;
  for (int i = 0; i < adj[strt].size(); i++) {
    if (adj[strt][i] != parent) {
      xr = (xr ^ dfs(adj, vis, val, adj[strt][i], strt));
    }
  }
  vis[strt] = (xr ^ vis[strt]);
  if (vis[strt] == target) {
    cnt++;
    vis[strt] = 0;
  }
  return vis[strt];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long term = (n * (n + 1)) / 2;
    vector<long long> a(n + 1);
    vector<long long> temp(n + 1);
    long long i, j, k, flag = 0, sum = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      sum += a[i];
      temp[i] = 0;
    }
    if (sum % term) {
      na();
      continue;
    }
    long long tt = sum / term;
    vector<long long> arr;
    for (i = 1; i < n; i++) {
      arr.push_back(a[i] - a[i - 1]);
    }
    arr.push_back(a[0] - a[n - 1]);
    for (i = 1; i <= n; i++) {
      j = i % n;
      temp[j] = tt - arr[i - 1];
      if (temp[j] % n) {
        flag = 1;
        break;
      }
      temp[j] /= n;
      if (temp[j] <= 0) {
        flag = 1;
        break;
      }
    }
    if (flag) {
      na();
    } else {
      ha();
      for (i = 0; i < n; i++) {
        cout << temp[i] << " ";
      }
      cout << endl;
    }
  }
  return 0;
}

```