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,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
//#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//------------------------------------------------------------------------------
//------------------------------------------------------------------------------
void _main()
{
    int N;
    cin >> N;
    vector<ll> A(N);
    ll sum = 0;
    rep(i, 0, N)
    {
        cin >> A[i];
        sum += A[i];
    }

    const bool ok = [&]()
    {
        const ll s = N * (N + 1) / 2;

        if (sum % s != 0)
        {
            return false;
        }

        const ll operationNum = sum / s;

        vector<ll> diff(N);
        rep(i, 0, N)
        {
            diff[i] = A[i] - A[(i - 1 + N) % N];
        }

        vector<ll> oneNum(N);
        rep(i, 0, N)
        {
            if ((operationNum - diff[i]) % N != 0)
            {
                return false;
            }

            oneNum[i] = (operationNum - diff[i]) / N;
        }

        ll oneNumSum = 0;
        rep(i, 0, N)
        {
            oneNumSum += oneNum[i];
        }

        return oneNumSum == operationNum;
    }();

    cout << (ok ? "YES" : "NO") << endl;
}