#include <bits/stdc++.h>
using namespace std;
bool dcmp(int x, int y) { return x > y; }
bool acmp(int x, int y) { return x < y; }
bool afcmp(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
bool dfcmp(pair<int, int> a, pair<int, int> b) { return a.first > b.first; }
bool ascmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
bool dscmp(pair<int, int> a, pair<int, int> b) { return a.second > b.second; }
int aabs(int a) { return a < 0 ? -a : a; }
long long tmax(long long a, long long b, long long c) {
  return max(max(a, b), c);
}
long long tmin(long long a, long long b, long long c) {
  return min(min(a, b), c);
}
bool in_range(int x, int l, int r) { return x >= l && x <= r; }
bool is_prime(int val) {
  if (val == 2) {
    return true;
  }
  if (val % 2 == 0) {
    return false;
  }
  for (int i = 3; i * i <= val; i += 2) {
    if (val % i == 0) {
      return false;
    }
  }
  return true;
}
int lowbit(int x) { return x & (-x); }
inline long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
const int maxN = 4e4 + 5;
bool finished = false;
int N, a[maxN];
long long ans[maxN];
void reset() {
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    cin >> a[i];
  }
}
void solve() {
  reset();
  if (finished) {
    return;
  }
  long long sum = 0, coeff = 0;
  for (int i = 1; i <= N; ++i) {
    sum += a[i];
    coeff += i;
  }
  if (sum % coeff != 0) {
    cout << "NO" << '\n';
    return;
  } else {
    sum /= coeff;
  }
  for (int i = 1; i <= N; ++i) {
    int diff = a[i] - a[i == 1 ? N : i - 1];
    long long x = sum - diff;
    if (x % N == 0 && x / N > 0) {
      ans[i] = x / N;
    } else {
      cout << "NO" << '\n';
      return;
    }
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= N; ++i) {
    cout << ans[i] << " ";
  }
  cout << '\n';
}
int main(void) {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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;
bool dcmp(int x, int y) { return x > y; }
bool acmp(int x, int y) { return x < y; }
bool afcmp(pair<int, int> a, pair<int, int> b) { return a.first < b.first; }
bool dfcmp(pair<int, int> a, pair<int, int> b) { return a.first > b.first; }
bool ascmp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
bool dscmp(pair<int, int> a, pair<int, int> b) { return a.second > b.second; }
int aabs(int a) { return a < 0 ? -a : a; }
long long tmax(long long a, long long b, long long c) {
  return max(max(a, b), c);
}
long long tmin(long long a, long long b, long long c) {
  return min(min(a, b), c);
}
bool in_range(int x, int l, int r) { return x >= l && x <= r; }
bool is_prime(int val) {
  if (val == 2) {
    return true;
  }
  if (val % 2 == 0) {
    return false;
  }
  for (int i = 3; i * i <= val; i += 2) {
    if (val % i == 0) {
      return false;
    }
  }
  return true;
}
int lowbit(int x) { return x & (-x); }
inline long long gcd(long long x, long long y) { return y ? gcd(y, x % y) : x; }
inline long long lcm(long long x, long long y) { return x / gcd(x, y) * y; }
const int maxN = 4e4 + 5;
bool finished = false;
int N, a[maxN];
long long ans[maxN];
void reset() {
  cin >> N;
  for (int i = 1; i <= N; ++i) {
    cin >> a[i];
  }
}
void solve() {
  reset();
  if (finished) {
    return;
  }
  long long sum = 0, coeff = 0;
  for (int i = 1; i <= N; ++i) {
    sum += a[i];
    coeff += i;
  }
  if (sum % coeff != 0) {
    cout << "NO" << '\n';
    return;
  } else {
    sum /= coeff;
  }
  for (int i = 1; i <= N; ++i) {
    int diff = a[i] - a[i == 1 ? N : i - 1];
    long long x = sum - diff;
    if (x % N == 0 && x / N > 0) {
      ans[i] = x / N;
    } else {
      cout << "NO" << '\n';
      return;
    }
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= N; ++i) {
    cout << ans[i] << " ";
  }
  cout << '\n';
}
int main(void) {
  std::ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
  return 0;
}

```