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 <bits/stdc++.h>
#include <unordered_set>
using namespace std;
#define ll long long
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define repeat(i,s,n) for(int (i)=s; (i)<(n); (i)++)
#define revrep(i,n) for(int (i)=(n)-1;i>=0; i--)

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<setprecision(std::numeric_limits<float>::max_digits10);
  int n;
  cin>>n;
  vector<ll> a(n);
  rep(i,n)cin>>a[i];
  ll sum=0;
  rep(i,n)sum+=a[i];
  if(sum%(n*(n+1)/2)!=0) {
    cout << "NO" << endl;
    return 0;
  }
  ll k = sum/(n*(n+1)/2);
  vector<ll> b(n);
  rep(i,n) b[i]=a[(i+1)%n]-a[i%n];
  ll c=0;
  rep(i,n) {
    b[i]-=k;
    if(b[i]>0||b[i]%n!=0) {
      cout << "NO" << endl;
      return 0;
    }
    c+=-b[i]/n;
  }
  if(c==k) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
  return 0;
}