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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
using namespace std;

int a[100005];
int d[100005];

long long int sumup(long long int n){
	return (n * (n + 1)) / 2;
}

int main(){
  int n;
  cin >> n;
  long long sum=0;
  for(int i=0;i<n;i++){ cin >> a[i]; sum+=a[i];}
  if(sum % sumup(n) != 0){
  	cout << "NO" << endl;
    return 0;
  }
  int t = sum / sumup(n);
  //d[i] denote a[i]-a[i-1];
  for(int i=0;i<n;i++) d[i] = a[(i+n) % n] - a[(i+n-1) % n] - t + 1;
  bool flag = true;
  for(int i=0;i<n;i++){
    if(d[i]>1){flag = false; break;}
    if(d[i]!=1){
      if((1 - d[i]) % n != 0){flag = false; break;}
      int tem = (1 - d[i]) / n;
      if(t < tem){flag = false; break;}
      d[i] = 1;
      t -= tem;
    }
  }
  if(t!=0||!flag) cout << "NO" << endl;
  else cout << "YES" << endl;
  return 0;
}