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: Wrong Answer)

#include <cstdio>
#include <cstring>
#include <cassert>
#include <iostream>
#include <algorithm>
#include <vector>
#include <set>

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR (i, 0, n)
#define _ << " _ " <<
#define TRACE(x) cerr << #x << " = " << x << endl
#define debug(...) fprintf(stderr, __VA_ARGS__)
//#define debug
//#define TRACE(x)

using namespace std;

typedef long long llint;

const int MAXN = 100010;

int n;
int a[MAXN];
llint sum;

int main(void) {
  scanf("%d",&n);
  REP(i, n) {
    scanf("%d",&a[i]);
    sum += a[i];
  }

  llint t = (llint)n*(n+1)/2;
  if (sum % t) {
    printf("NO\n");
    return 0;
  }
  llint k = sum / t;

  a[n] = a[0];
  REP(i, n) {
    llint x = (a[i+1] - a[i]) % n;
    if (x < 0) x += n;
    if (x != k) {
      printf("NO\n");
      return 0;
    }
  }

  printf("YES\n");
    
  return 0;
}