#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
template <typename T>
void printv(const T& t) {
  std::copy(t.cbegin(), t.cend(),
            std::ostream_iterator<typename T::value_type>(std::cout, ", "));
  cout << '\n';
}
long long int modpower(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    res %= mod;
    a = a * a;
    a %= mod;
  }
  return res;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long int expo(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long int mod_inv(long long int a, long long int b) {
  return expo(a, b - 2, b);
}
long long int mod_add(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long int mod_mul(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long int mod_sub(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mod_inv(b, m), m) + m) % m;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  long long int sum = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    sum += v[i];
  }
  long long int x = n * (n + 1);
  x /= 2;
  if (sum % x != 0) {
    cout << "NO\n";
    return;
  }
  long long int s = sum / x;
  vector<long long int> ans(n);
  for (long long int i = 1; i < n; i++) {
    long long int curr = s - (v[i] - v[i - 1]);
    if (curr <= 0 || curr % n != 0) {
      cout << "NO\n";
      return;
    }
    curr /= n;
    ans[i] = curr;
  }
  long long int curr = s - (v[0] - v[n - 1]);
  if (curr <= 0 || curr % n != 0) {
    cout << "NO\n";
    return;
  }
  curr /= n;
  ans[0] = curr;
  cout << "YES\n";
  for (long long int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
  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;
#pragma GCC optimize("O3")
#pragma GCC target("avx2")
template <typename T>
void printv(const T& t) {
  std::copy(t.cbegin(), t.cend(),
            std::ostream_iterator<typename T::value_type>(std::cout, ", "));
  cout << '\n';
}
long long int modpower(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a;
    res %= mod;
    a = a * a;
    a %= mod;
  }
  return res;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
long long int expo(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long int mod_inv(long long int a, long long int b) {
  return expo(a, b - 2, b);
}
long long int mod_add(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long int mod_mul(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long int mod_sub(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mod_inv(b, m), m) + m) % m;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> v(n);
  long long int sum = 0;
  for (long long int i = 0; i < n; i++) {
    cin >> v[i];
    sum += v[i];
  }
  long long int x = n * (n + 1);
  x /= 2;
  if (sum % x != 0) {
    cout << "NO\n";
    return;
  }
  long long int s = sum / x;
  vector<long long int> ans(n);
  for (long long int i = 1; i < n; i++) {
    long long int curr = s - (v[i] - v[i - 1]);
    if (curr <= 0 || curr % n != 0) {
      cout << "NO\n";
      return;
    }
    curr /= n;
    ans[i] = curr;
  }
  long long int curr = s - (v[0] - v[n - 1]);
  if (curr <= 0 || curr % n != 0) {
    cout << "NO\n";
    return;
  }
  curr /= n;
  ans[0] = curr;
  cout << "YES\n";
  for (long long int i = 0; i < n; i++) cout << ans[i] << " ";
  cout << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```