#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<ld, ld>;
template <class T>
using V = vector<T>;
using vi = V<int>;
using vb = V<bool>;
using vl = V<ll>;
using vd = V<ld>;
using vs = V<str>;
using vpi = V<pi>;
using vpl = V<pl>;
using vpd = V<pd>;
template <class T>
int lwb(V<T>& a, const T& b) {
  return int(lower_bound(begin(a), end(a), b) - bg(a));
}
template <class T>
int upb(V<T>& a, const T& b) {
  return int(upper_bound(begin(a), end(a), b) - bg(a));
}
const int MOD = 1e9 + 7;
const int MX = 1e5 + 5;
const ll BIG = 1e18;
const int dx[4]{1, 0, -1, 0}, dy[4]{0, 1, 0, -1};
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
constexpr int p2(int x) { return 1 << x; }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
ll gcd(T& a, T& b) {
  if (a > b) swap(a, b);
  while (a) {
    b -= a;
    if (a > b) swap(a, b);
  }
  return b;
}
template <class T>
ll power(T& a, T& b) {
  ll ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    a *= a;
    b >>= 1;
  }
  return ans;
}
void solve() {
  int n;
  cin >> n;
  int b[n], a[n];
  ll sum = 0;
  for (int i = (0); i < (n); ++i) {
    cin >> b[i];
    sum += b[i];
  }
  ll cnt = n * 1ll * (n + 1) / 2;
  if (sum % cnt != 0) {
    cout << "NO\n";
    return;
  }
  for (int i = (n)-1; i >= (0); --i) {
    ll res = b[i] - b[(i + 1) % n] + (sum / cnt);
    if (res % n != 0 || res <= 0) {
      cout << "NO\n";
      return;
    }
    a[(i + 1) % n] = res / n;
  }
  cout << "YES\n";
  for (int i = (0); i < (n); ++i) cout << a[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

### Prompt

Please create a solution in Cpp 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;
using ll = long long;
using ld = long double;
using str = string;
using pi = pair<int, int>;
using pl = pair<ll, ll>;
using pd = pair<ld, ld>;
template <class T>
using V = vector<T>;
using vi = V<int>;
using vb = V<bool>;
using vl = V<ll>;
using vd = V<ld>;
using vs = V<str>;
using vpi = V<pi>;
using vpl = V<pl>;
using vpd = V<pd>;
template <class T>
int lwb(V<T>& a, const T& b) {
  return int(lower_bound(begin(a), end(a), b) - bg(a));
}
template <class T>
int upb(V<T>& a, const T& b) {
  return int(upper_bound(begin(a), end(a), b) - bg(a));
}
const int MOD = 1e9 + 7;
const int MX = 1e5 + 5;
const ll BIG = 1e18;
const int dx[4]{1, 0, -1, 0}, dy[4]{0, 1, 0, -1};
mt19937 rng((uint32_t)chrono::steady_clock::now().time_since_epoch().count());
template <class T>
using pqg = priority_queue<T, vector<T>, greater<T>>;
constexpr int pct(int x) { return __builtin_popcount(x); }
constexpr int bits(int x) { return x == 0 ? 0 : 31 - __builtin_clz(x); }
constexpr int p2(int x) { return 1 << x; }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); }
ll fdiv(ll a, ll b) { return a / b - ((a ^ b) < 0 && a % b); }
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
template <class T>
ll gcd(T& a, T& b) {
  if (a > b) swap(a, b);
  while (a) {
    b -= a;
    if (a > b) swap(a, b);
  }
  return b;
}
template <class T>
ll power(T& a, T& b) {
  ll ans = 1;
  while (b) {
    if (b & 1) ans *= a;
    a *= a;
    b >>= 1;
  }
  return ans;
}
void solve() {
  int n;
  cin >> n;
  int b[n], a[n];
  ll sum = 0;
  for (int i = (0); i < (n); ++i) {
    cin >> b[i];
    sum += b[i];
  }
  ll cnt = n * 1ll * (n + 1) / 2;
  if (sum % cnt != 0) {
    cout << "NO\n";
    return;
  }
  for (int i = (n)-1; i >= (0); --i) {
    ll res = b[i] - b[(i + 1) % n] + (sum / cnt);
    if (res % n != 0 || res <= 0) {
      cout << "NO\n";
      return;
    }
    a[(i + 1) % n] = res / n;
  }
  cout << "YES\n";
  for (int i = (0); i < (n); ++i) cout << a[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
}

```