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 <limits.h>
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <utility>
#include <vector>

#define int long long
#define REP(i, n) for (int i = 0, i##_len = (n); i < i##_len; ++i)
#define REPS(i, n) for (int i = 1, i##_len = (n); i <= i##_len; ++i)
#define FOR(i, a, b) for (int i = (a), i##_len = (b); i <= i##_len; ++i)
#define REV(i, a, b) for (int i = (a); i >= (b); --i)
#define CLR(a, b) memset((a), (b), sizeof(a))
#define DUMP(x) cout << #x << " = " << (x) << endl;
#define INF (LLONG_MAX - 1e5)

using namespace std;

int N, A[100005];
int a, na, b, nb;

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    cin >> N;
    REP(i, N) cin >> A[i];

    vector<int> v;
    map<int, int> m;
    REP(i, N) {
        int next = i == N - 1 ? 0 : i + 1;
        v.push_back(A[next] - A[i]);
        m[A[next] - A[i]]++;
    }

    if (m.size() > 2) {
        cout << "NO" << endl;
        return 0;
    }

    int t = -1;
    int changed = 0;
    int prev = v[0];
    REP(i, N) {
        if (v[i] != prev) {
            changed++;
        }
        if (v[i] < 0) t = i;
        prev = v[i];
    }
    if (changed > 2) {
        cout << "NO" << endl;
        return 0;
    }
    if (changed == 1) t = N - 1;
    for (auto kv : m) {
        if (kv.first >= 0) {
            a = kv.first;
            na = kv.second;
        } else {
            b = kv.first;
            nb = kv.second;
        }
    }

    if (a - 1 != nb - 1) {
        cout << "NO" << endl;
        return 0;
    }
    if (b + -1 * (nb - 1) != 1 - N) {
        cout << "NO" << endl;
        return 0;
    }

    int k = nb - 1;
    while (k) {
        A[t] -= k;
        k--;
    }

    if (A[t] != N) {
        cout << "NO" << endl;
        return 0;
    }


    cout << "YES" << endl;

    return 0;
}