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;
#define FOR(i,a,b) for(int i =(a);i<(b);i++)
#define REP(i,n) for(int i=0;i<(n);i++)
#define RREP(i,n) for(int i=(n)-1;i>=0;i--)
#define REPS(i,n) for(int i=1;i<=(n);i++)
#define RREPS(i,n) for(int i=(n);i>0;i--)
#define mp make_pair
#define pb(a) push_back(a)
typedef long long ll;
ll d[100000];
ll A[100000];
int main(){
  ll N;
  cin >> N;
  ll sum = 0;
  REP(i,N){
    cin >> A[i];
    sum += A[i];
  }
  if(sum%(N*(N+1)/2) != 0){
    cout << "NO" << endl;
    return 0;
  }
  ll k = sum/(N*(N+1)/2);
  d[0] = A[0] - A[N-1] - k;
  REPS(i,N-1) d[i] = A[i] - A[i-1] - k;
  bool ans = true;
  REP(i,N){
    if(d[i] > 0 || d[i]%N != 0){
      ans = false;
      break;
    }
  }
  if(ans) cout << "YES" << endl;
  else cout << "NO" << endl;
  return 0;
}