Problem: While discussing a proper problem A for a Codeforces Round, Kostya created a cyclic array of positive integers a_1, a_2, …, a_n. Since the talk was long and not promising, Kostya created a new cyclic array b_1, b_2, …, b_{n} so that b_i = (a_i mod a_{i + 1}), where we take a_{n+1} = a_1. Here mod is the [modulo operation](https://en.wikipedia.org/wiki/Modulo_operation). When the talk became interesting, Kostya completely forgot how array a had looked like. Suddenly, he thought that restoring array a from array b would be an interesting problem (unfortunately, not A).

Input

The first line contains a single integer n (2 ≤ n ≤ 140582) — the length of the array a.

The second line contains n integers b_1, b_2, …, b_{n} (0 ≤ b_i ≤ 187126).

Output

If it is possible to restore some array a of length n so that b_i = a_i mod a_{(i mod n) + 1} holds for all i = 1, 2, …, n, print «YES» in the first line and the integers a_1, a_2, …, a_n in the second line. All a_i should satisfy 1 ≤ a_i ≤ 10^{18}. We can show that if an answer exists, then an answer with such constraint exists as well.

It it impossible to restore any valid a, print «NO» in one line.

You can print each letter in any case (upper or lower).

Examples

Input

4
1 3 1 0


Output

YES
1 3 5 2


Input

2
4 4


Output

NO

Note

In the first example:

  * 1 mod 3 = 1 
  * 3 mod 5 = 3 
  * 5 mod 2 = 1 
  * 2 mod 1 = 0 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long MOD = 1000000000000000000;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long n, m;
  cin >> n;
  long long b[n];
  long long mx = 0;
  for (int i = 0; i < n; i++) {
    cin >> b[i];
    mx = max(mx, b[i]);
  }
  long long x = -1;
  for (int i = 0; i < n; i++) {
    if (b[(i + n - 1) % n] < b[i]) {
      x = i;
    }
  }
  if (x == -1) {
    if (mx > 0) {
      cout << "NO";
      return 0;
    } else {
      cout << "YES" << endl;
      for (int i = 0; i < n; i++) {
        cout << "1 ";
      }
      return 0;
    }
  }
  long long a[n];
  a[x] = b[x];
  a[(x - 1 + n) % n] = b[x] * (mx + 1000) + b[(x - 1 + n) % n];
  long long t = (x - 2 + n) % n;
  while (t != x) {
    a[t] = a[(t + 1) % n] + b[t];
    t = (t - 1 + n) % n;
  }
  cout << "YES" << endl;
  for (auto i : a) {
    cout << i << " ";
  }
}