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;

int main(void) {
    long n;
    cin >> n;
    vector<long> a(n);
    for(auto &e: a) cin >> e;

    long sum = 0;
    for(auto &e: a) sum += e;
    long cnt = sum / (n * (n+1) /2);

    if(cnt * (n*(n+1)/2) != sum) {
        cout << "NO" << endl;
        return 0;
    }

    long t = 0;
    for(long i=0; i<n; i++) {
        // a[i+1] - a[i] == x - (n-1)*(cnt-x)
        // a[i+1] - a[i] == x - (n-1)*cnt + (n-1)x
        // nx == a[i+1] - a[i] + (n-1)*cnt
        long nx = a[(i+1)%n] - a[i] + (n-1)*cnt;
        long x = nx / n;
        if (x*n != nx) {
            cout << "NO" << endl;
            return 0;
        }
        t += cnt - x;
    }

    if(t != cnt) {
        cout << "NO" << endl;
        return 0;
    }

    cout << "YES" << endl;
    return 0;
}