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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main(){
	int N;
	int i,flag=0;
	int max=0,maxi=0,hiku=0,sum=0,min;
	long long int A[100001];
	scanf("%d",&N);
	for(i=0;i<N;i++){
		scanf("%lld",&A[i]);
	}
	min=A[0];
	for(i=1;i<=N;i++){
		sum+=i;
	}
	for(i=1;i<=N;i++){
		if(min>A[i]) min=A[i];
	}
	
	if(min>=sum) {
		for(i=0;i<N;i++){
			A[i]-=sum*(min/sum);
		}
	}
	
	
	while(1){
		max=0;
		for(i=0;i<N;i++){
			if(A[i]<0){
				flag=2;
				break;
			}
			if(max<A[i]){
			max=A[i];
			maxi=i;
			}
		}
		if(max<N || flag==2) break;
		for(i=1;i<=N;i++){
			A[(maxi+i)%N]-=i;
			}
	}
	for(i=0;i<N;i++){
		if(A[i]!=0){
		flag=1;
		break;
		}
	}
	if(flag==0) printf("YES\n");
	else printf("NO\n");
	
	return 0;
}