description: 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. 
solution: #include <bits/stdc++.h>
  using namespace std;
  template <typename T>
  inline bool read(T &x) {
    x = 0;
    int f = 1;
    char c = getchar();
    for (; c != EOF && !isdigit(c); c = getchar())
      if (c == '-') f = -1;
    if (c == EOF) return 0;
    for (; c != EOF && isdigit(c); c = getchar())
      x = (x << 1) + (x << 3) + (c ^ 48);
    x *= f;
    return 1;
  }
  template <typename T>
  void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) print(x / 10);
    putchar('0' + x % 10);
  }
  const int N = 5e4;
  long long T, n, b[N], g[N], sum, ans[N], mn, t;
  int main() {
    read(T);
    while (T--) {
      read(n);
      sum = 0;
      mn = 1e12;
      for (int i = 1; i <= n; ++i) read(b[i]);
      for (int i = 2; i <= n; ++i) sum += (g[i] = b[i] - b[i - 1]);
      sum += (g[1] = b[1] - b[n]);
      if (n == 1) {
        printf("YES\n%lld\n", b[1]);
        continue;
      }
      if (sum) goto FAIL;
      for (int i = 2; i <= n; ++i)
        if ((g[1] - g[i]) % n)
          goto FAIL;
        else
          ans[i] = (g[1] - g[i]) / n, mn = min(ans[i], mn);
      mn = max(1ll, -mn);
      sum = 0;
      t = (n * (n + 1)) / 2;
      for (int i = 2; i <= n; ++i) sum += ans[i] * (n + 2 - i);
      if ((b[1] - sum) <= 0 || (b[1] - sum) % t ||
          (ans[1] = (b[1] - sum) / t) <= -mn)
        goto FAIL;
      if ((ans[1] = (b[1] - sum) / t) >= b[1]) goto FAIL;
      for (int i = 2; i <= n; ++i)
        if ((ans[i] += ans[1]) >= b[i] || !ans[i]) goto FAIL;
      puts("YES");
      for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
      puts("");
      continue;
    FAIL:
      puts("NO");
    }
    return 0;
  }
  