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: Accepted)

#include <bits/stdc++.h>

using namespace std;

const int MAXN = 100000 + 10;

int n;
int a[MAXN], b[MAXN];

string solve() {
    cin >> n;
    long long suma = 0;
    long long x = 0;
    for(int i = 1; i <= n; ++i) {
        cin >> a[i];
        suma += a[i];
        x += i;
    }
    if (suma % x != 0) return "NO";
    int k = suma / x;
    for(int i = 1; i <= n; ++i) {
        int j = i + 1;
        if (j > n) j = 1;
        long long v = (a[i] + k - a[j]);
        if (v % n != 0) return "NO";
        if (v < 0) return "NO";
        b[i] = v / n;
    }
    return "YES";
}

int main() {
    int ntests = 1;
    //cin >> ntests;
    for(int tc = 1; tc <= ntests; ++tc) {
        cout << solve() << "\n";
    }
}