#include <bits/stdc++.h>
using namespace std;
constexpr long long int mod = 1e9 + 7;
auto seed = chrono::high_resolution_clock::now().time_since_epoch().count();
std::mt19937 mt(seed);
int myrand(int mod) { return (int)((((double)rand() / RAND_MAX) * (mod))); }
template <typename T, typename U>
void operator%=(T &a, const U &m) {
  if (abs(a) >= m) a %= m;
  if (a < 0) a += m;
}
template <typename T, typename U>
T operator%(T &a, const U &m) {
  T b = a;
  b %= m;
  return b;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (int i = int(0); i < int((int)a.size()); i++) out << a[i] << " ";
  out << "\n";
  return out;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &a) {
  for (int i = int(0); i < int((int)a.size()); i++) in >> a[i];
  return in;
}
template <typename T, typename U>
T pow2(const T &x, const U &y) {
  T res = 1, X = x;
  U Y = y;
  while (Y > 0) {
    if (Y & 1) res *= X;
    Y >>= 1;
    X *= X;
    X %= mod;
    res %= mod;
  }
  return res;
}
pair<long long int, long long int> operator-(
    pair<long long int, long long int> &a,
    pair<long long int, long long int> &b) {
  return {a.first - b.first, a.second - b.second};
}
pair<long long int, long long int> operator+(
    pair<long long int, long long int> &a,
    pair<long long int, long long int> &b) {
  return {a.first + b.first, a.second + b.second};
}
void solve() {
  int n;
  cin >> n;
  vector<long long int> a(n);
  cin >> a;
  long long int N = n * (n + 1) / 2;
  long long int sum = accumulate(a.begin(), a.end(), 0LL);
  if (sum % N != 0) {
    cout << "NO\n";
    return;
  }
  sum /= N;
  vector<long long int> res(n);
  for (int i = 0; i < n; i++) {
    int j = (i + 1) % n;
    long long int c = a[j] - a[i];
    res[j] = sum - c;
    if (res[j] <= 0 || res[j] % n != 0) {
      cout << "NO\n";
      return;
    } else {
      res[j] /= n;
    }
  }
  cout << "YES\n";
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Your task is to create 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;
constexpr long long int mod = 1e9 + 7;
auto seed = chrono::high_resolution_clock::now().time_since_epoch().count();
std::mt19937 mt(seed);
int myrand(int mod) { return (int)((((double)rand() / RAND_MAX) * (mod))); }
template <typename T, typename U>
void operator%=(T &a, const U &m) {
  if (abs(a) >= m) a %= m;
  if (a < 0) a += m;
}
template <typename T, typename U>
T operator%(T &a, const U &m) {
  T b = a;
  b %= m;
  return b;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &a) {
  for (int i = int(0); i < int((int)a.size()); i++) out << a[i] << " ";
  out << "\n";
  return out;
}
template <typename T>
istream &operator>>(istream &in, vector<T> &a) {
  for (int i = int(0); i < int((int)a.size()); i++) in >> a[i];
  return in;
}
template <typename T, typename U>
T pow2(const T &x, const U &y) {
  T res = 1, X = x;
  U Y = y;
  while (Y > 0) {
    if (Y & 1) res *= X;
    Y >>= 1;
    X *= X;
    X %= mod;
    res %= mod;
  }
  return res;
}
pair<long long int, long long int> operator-(
    pair<long long int, long long int> &a,
    pair<long long int, long long int> &b) {
  return {a.first - b.first, a.second - b.second};
}
pair<long long int, long long int> operator+(
    pair<long long int, long long int> &a,
    pair<long long int, long long int> &b) {
  return {a.first + b.first, a.second + b.second};
}
void solve() {
  int n;
  cin >> n;
  vector<long long int> a(n);
  cin >> a;
  long long int N = n * (n + 1) / 2;
  long long int sum = accumulate(a.begin(), a.end(), 0LL);
  if (sum % N != 0) {
    cout << "NO\n";
    return;
  }
  sum /= N;
  vector<long long int> res(n);
  for (int i = 0; i < n; i++) {
    int j = (i + 1) % n;
    long long int c = a[j] - a[i];
    res[j] = sum - c;
    if (res[j] <= 0 || res[j] % n != 0) {
      cout << "NO\n";
      return;
    } else {
      res[j] /= n;
    }
  }
  cout << "YES\n";
  cout << res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```