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 <cmath>
#include <queue>
#include <vector>
#include <time.h>
#include <string>
#include <stack>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int, int> Pii;
const int inf = 0x3f3f3f3f;
const LL INF = (1uLL << 63) - 1;
const LL mod = 1000000007;
const int N = 150 + 5;
const double Pi = acos(-1.0);
const int maxn = 2e5 + 5;
LL a[maxn];
int main() {
#ifdef local
    freopen("in", "r", stdin);
    //  freopen("w","w",stdout);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin>>n;
    LL del = 0;
    LL d = (1 + n) * n / 2;
    for(int i = 0; i < n; i++){
        cin>>a[i];
        del += a[i];
    }
    if(del % d != 0){
        cout<<"NO"<<endl;
    }
    else {
        LL time = del / d;
        LL cut = 0;
        bool flag = 0;
        for(int i = 1; i < n; i++){
            cut += a[i] - a[i - 1];
            if(a[i] - a[i-1] > time||a[i] - a[i-1] < time * (1 - n)){
                flag = 1;
            }
        }
        if((cut + time) % n == 0 && (cut + time) / n <= time && !flag){
            cout<<"YES"<<endl;
        }
        else cout<<"NO"<<endl;
    }
}