Problem:
Score : 500 points

Problem Statement

There are N boxes arranged in a circle. The i-th box contains A_i stones.

Determine whether it is possible to remove all the stones from the boxes by repeatedly performing the following operation:

Select one box. Let the box be the i-th box. Then, for each j from 1 through N, remove exactly j stones from the (i+j)-th box. Here, the (N+k)-th box is identified with the k-th box.

Note that the operation cannot be performed if there is a box that does not contain enough number of stones to be removed.

Constraints

1 ≦ N ≦ 10^5

1 ≦ A_i ≦ 10^9

Input

The input is given from Standard Input in the following format:

N
A_1 A_2 … A_N

Output

If it is possible to remove all the stones from the boxes, print YES. Otherwise, print NO.

Sample Input 1

5
4 5 1 2 3

Sample Output 1

YES

All the stones can be removed in one operation by selecting the second box.

Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

Sample Input 3

4
1 2 3 1

Sample Output 3

NO

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#define finish(x) return cout << x << endl, 0
#define ll long long

int n;
vector <int> a;

int main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n;
    a.resize(n);
    ll sum = 0;
    for(auto &i : a){
        cin >> i;
        sum += i;
    }
    ll s = 1LL * n * (n + 1) / 2;
    if(sum % s) finish("NO");
    ll h = sum / s;
    ll cnt = 0;
    for(int i = 0 ; i < n ; i++){
        ll d = a[i] - a[(i - 1 + n) % n] - h;
        if(d % n) finish("NO");
        cnt += d / n;
    }
    if(-cnt == h) cout << "YES\n";
    else cout << "NO\n";
}