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: Accepted)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) x.begin(),x.end()
const long double PI = acos(-1.0L);
const long long MOD = 1000000007LL;
//const long long MOD = 998244353LL;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true;} return false; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true;} return false; }
///////////////////////////////////////////////////////////////////////////////////////////////////

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    ll N; cin >> N;
    vector<ll> A(N);
    rep(i,N) cin >> A[i];

    ll M = N*(N+1)/2;
    ll S = accumulate(all(A),0LL);
    if (S%M != 0) {
        cout << "NO" << endl;
        return 0;
    }
    ll K = S/M;
    vector<ll> cnt(N);
    rep(i,N-1) {
        if ((A[i]+K-A[i+1])%N) {
            cout << "NO" << endl;
            return 0;
        }
        cnt[i+1] = (A[i]+K-A[i+1])/N;
        A[i+1] = A[i]+K;
        if (cnt[i+1] < cnt[i]) {
            cout << "NO" << endl;
            return 0;    
        }
    }
    ll z = K - cnt[N-1];
    if (z < 0) {
        cout << "NO" << endl;
        return 0;
    }
    for (int i = N-1; i > 0; i--) {
        cnt[i] -= cnt[i-1];
    }
    cnt[0] = z;
    ll c = 0;
    vector<ll> B(N);
    rep(i,N) B[i] = N+1-i;
    B[0] -= N;
    rep(i,N) {
        c += cnt[i]*B[i];
    }
    cout << (c == A[0] ? "YES" : "NO") << endl;
}