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>
using namespace std;

typedef long long int ll;
int n;

bool dfs(vector<ll> v) {

    bool ans = false;

    ll t = 1LL<<32;
    ll m = -1LL;
    for (int i = 0; i < n; i++) {
        t = min(t, v[i]);
        m = max(m, v[i]);
    }

    if (!m) return true;

    set<int> s;
    for (int i = 0; i < n; i++) if (v[i] == t) s.insert(i);

    for (auto j : s) {
        auto vv = v;
        ll l = 1LL<<32;
        for (int i = 0; i < n; i++) {
            l = min(l, v[(i+j)%n] / (ll)(i+1));
        }

        if (!l) continue;

        for (int i = 0; i < n; i++) {
            vv[(i+j)%n] -= (ll)(i+1) * l;
        }

        if (dfs(vv)) ans = true;
    }

    return ans;
}

int main() {
    cin >> n;
    vector<ll> v(n);
    for (auto& i : v) cin >> i;

    if (dfs(v))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;

    return 0;
}