#include <bits/stdc++.h>
using namespace std;
const long long mod1 = 1000000007;
const long long mod2 = 998244353;
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result *= a;
    }
    a *= a;
    b /= 2;
  }
  return result;
}
long long gcd(long long x, long long y) {
  long long r;
  while (y != 0 && (r = x % y) != 0) {
    x = y;
    y = r;
  }
  return y == 0 ? x : y;
}
long long countSetBits(long long x) {
  long long Count = 0;
  while (x > 0) {
    if (x & 1) Count++;
    x = x >> 1;
  }
  return Count;
}
long long mod(long long x, long long M) { return ((x % M + M) % M); }
long long add(long long a, long long b, long long M) {
  return mod(mod(a, M) + mod(b, M), M);
}
long long mul(long long a, long long b, long long M) {
  return mod(mod(a, M) * mod(b, M), M);
}
long long powerM(long long a, long long b, long long M) {
  long long res = 1ll;
  while (b) {
    if (b % 2ll == 1ll) {
      res = mul(a, res, M);
    }
    a = mul(a, a, M);
    b /= 2ll;
  }
  return res;
}
long long mod_inv(long long a, long long m) {
  long long g = m, r = a, x = 0, y = 1;
  while (r != 0) {
    long long q = g / r;
    g %= r;
    swap(g, r);
    x -= q * y;
    swap(x, y);
  }
  return mod(x, m);
}
long long ncr(long long n, long long k) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
vector<long long> FA, INV;
void fact(long long n, long long M) {
  FA.resize(n + 1);
  FA[0] = 1;
  for (long long i = 1; i <= n; i++) {
    FA[i] = (FA[i - 1] * i) % M;
  }
}
void icalc(long long n, long long M) {
  INV.resize(n + 1);
  INV[n] = mod_inv(FA[n], M);
  for (long long i = n - 1; i >= 0; i--) {
    INV[i] = (INV[i + 1] * (i + 1)) % M;
  }
}
long long ncrm(long long n, long long r, long long M) {
  if (n < r) return 0;
  if (r == 0) return 1;
  long long a = FA[n];
  a *= INV[r];
  a %= M;
  a *= INV[n - r];
  a %= M;
  return a;
}
long long rand_int(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
ifstream fin;
ofstream fout;
template <typename T>
ostream& operator<<(ostream& output, const vector<T>& data) {
  for (const T& x : data) output << x << " ";
  return output;
}
template <typename T>
istream& operator>>(istream& input, vector<T>& data) {
  for (auto& item : data) {
    input >> item;
  }
  return input;
}
void solve() {
  int n;
  cin >> n;
  vector<int> ar(n);
  for (auto& x : ar) {
    cin >> x;
  }
  long long sum = accumulate(ar.begin(), ar.end(), 0LL);
  if ((sum) % ((n * 1LL * (n + 1)) / 2)) {
    cout << "NO\n";
    return;
  }
  sum /= (n * 1LL * (n + 1)) / 2;
  bool check = true;
  vector<int> ans(n);
  for (int i = 0; i < n; ++i) {
    int x = ar[(i + 1) % n] - ar[i];
    ans[(i + 1) % n] = (sum - x) / n;
    if ((sum - x) % n) {
      check = false;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (ans[i] <= 0) {
      check = false;
      break;
    }
  }
  if (check) {
    cout << "YES\n";
    for (auto it : ans) {
      cout << it << " ";
    }
  } else {
    cout << "NO";
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(20);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const long long mod1 = 1000000007;
const long long mod2 = 998244353;
mt19937 rng(
    (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
long long power(long long a, long long b) {
  long long result = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      result *= a;
    }
    a *= a;
    b /= 2;
  }
  return result;
}
long long gcd(long long x, long long y) {
  long long r;
  while (y != 0 && (r = x % y) != 0) {
    x = y;
    y = r;
  }
  return y == 0 ? x : y;
}
long long countSetBits(long long x) {
  long long Count = 0;
  while (x > 0) {
    if (x & 1) Count++;
    x = x >> 1;
  }
  return Count;
}
long long mod(long long x, long long M) { return ((x % M + M) % M); }
long long add(long long a, long long b, long long M) {
  return mod(mod(a, M) + mod(b, M), M);
}
long long mul(long long a, long long b, long long M) {
  return mod(mod(a, M) * mod(b, M), M);
}
long long powerM(long long a, long long b, long long M) {
  long long res = 1ll;
  while (b) {
    if (b % 2ll == 1ll) {
      res = mul(a, res, M);
    }
    a = mul(a, a, M);
    b /= 2ll;
  }
  return res;
}
long long mod_inv(long long a, long long m) {
  long long g = m, r = a, x = 0, y = 1;
  while (r != 0) {
    long long q = g / r;
    g %= r;
    swap(g, r);
    x -= q * y;
    swap(x, y);
  }
  return mod(x, m);
}
long long ncr(long long n, long long k) {
  if (n < k) return 0;
  if (k == 0) return 1;
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
vector<long long> FA, INV;
void fact(long long n, long long M) {
  FA.resize(n + 1);
  FA[0] = 1;
  for (long long i = 1; i <= n; i++) {
    FA[i] = (FA[i - 1] * i) % M;
  }
}
void icalc(long long n, long long M) {
  INV.resize(n + 1);
  INV[n] = mod_inv(FA[n], M);
  for (long long i = n - 1; i >= 0; i--) {
    INV[i] = (INV[i + 1] * (i + 1)) % M;
  }
}
long long ncrm(long long n, long long r, long long M) {
  if (n < r) return 0;
  if (r == 0) return 1;
  long long a = FA[n];
  a *= INV[r];
  a %= M;
  a *= INV[n - r];
  a %= M;
  return a;
}
long long rand_int(long long l, long long r) {
  return uniform_int_distribution<long long>(l, r)(rng);
}
ifstream fin;
ofstream fout;
template <typename T>
ostream& operator<<(ostream& output, const vector<T>& data) {
  for (const T& x : data) output << x << " ";
  return output;
}
template <typename T>
istream& operator>>(istream& input, vector<T>& data) {
  for (auto& item : data) {
    input >> item;
  }
  return input;
}
void solve() {
  int n;
  cin >> n;
  vector<int> ar(n);
  for (auto& x : ar) {
    cin >> x;
  }
  long long sum = accumulate(ar.begin(), ar.end(), 0LL);
  if ((sum) % ((n * 1LL * (n + 1)) / 2)) {
    cout << "NO\n";
    return;
  }
  sum /= (n * 1LL * (n + 1)) / 2;
  bool check = true;
  vector<int> ans(n);
  for (int i = 0; i < n; ++i) {
    int x = ar[(i + 1) % n] - ar[i];
    ans[(i + 1) % n] = (sum - x) / n;
    if ((sum - x) % n) {
      check = false;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (ans[i] <= 0) {
      check = false;
      break;
    }
  }
  if (check) {
    cout << "YES\n";
    for (auto it : ans) {
      cout << it << " ";
    }
  } else {
    cout << "NO";
  }
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(20);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```