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 <iostream>
#include <cstdio>
#include <algorithm>
#include <vector>
using namespace std;
#define SZ(a) (int)(a).size()
typedef long long ll;

int main() {
	cin.sync_with_stdio(false);
	int n;
	cin >> n;
	vector<int> a(n);
	auto sum = 0ll;
	for (auto i = 0; i < n; i++) {
		int x;
		cin >> x;
		sum += x;
		a[i] = x;
	}
	auto n2 = (ll)n*(n+1)/2;
	if (sum%n2 != 0) {
		puts("NO");
		return 0;
	}
	auto m = sum/n2;
	auto k = 0ll;
	for (auto i = 0; i < n; i++) {
		auto x = a[i]+m-a[(i+1)%n];
		if (x < 0 || x%n != 0) {
			puts("NO");
			return 0;
		}
		auto y = x/n;
		k += y;
	}
	if (k != m) {
		puts("NO");
		return 0;
	}
	puts("YES");
	return 0;
}