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>
#define REP(i, n) for(int i = 0;i < n;i++)
#define REPR(i, n) for(int i = n;i >= 0;i--)
#define FOR(i, m, n) for(int i = m;i < n;i++)
#define FORR(i, m, n) for(int i = m;i >= n;i--)
#define SORT(v, n) sort(v, v+n);
#define SORTR(v,n) sort(v, v+n, greater<int>());
#define ll long long
#define int long long
#define pb push_back
#define INF 100000000
using namespace std;
typedef pair<int, int> P; // cannot use P as var
typedef pair<ll, ll> LP;
typedef pair<int, P> PP;
typedef pair<ll, LP> LPP;

const int MAX_n = 100000;
int n,a[MAX_n];


signed main(){

    scanf("%lld", &n);
    REP(i,n){
        scanf("%lld", &a[i]);
    }
    //超賢い．解説見る．
    int sum=0;
    REP(i,n){
        sum+=a[i];
    }
    ll k = (ll)n * (n + 1) / 2;
    if(sum%k!=0){
        printf("NO\n");
        return 0;
    }

    k=sum/k;

    int d[MAX_n];
    REP(i,n-1){
        d[i]=a[i+1]-a[i];
    }
    d[n-1]=a[0]-a[n-1];

    REP(i,n){
        d[i]-=k;
    }

    REP(i,k){
        REP(i,n){
            if(d[i]!=0){
                d[i]+=5;
                break;
            }
        }
    }

    int check;
    REP(i,n){
        check++;
        if(d[i] != 0){
            break;
        }
    }

    if(check == n) printf("YES\n");
    else printf("NO\n");



    return 0;
}