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)

// Copyright (C) 2017 __debug.

// This program is free software; you can redistribute it and/or
// modify it under the terms of the GNU General Public License as
// published by the Free Software Foundation; version 3

// This program is distributed in the hope that it will be useful,
// but WITHOUT ANY WARRANTY; without even the implied warranty of
// MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
// GNU General Public License for more details.

// You should have received a copy of the GNU General Public License
// along with this program; If not, see <http://www.gnu.org/licenses/>.


#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/priority_queue.hpp>

#define x first
#define y second
#define MP std::make_pair
#define SZ(x) ((int)(x).size())
#define ALL(x) (x).begin(), (x).end()
#define DEBUG(...) fprintf(stderr, __VA_ARGS__)
#ifdef __linux__
#define getchar getchar_unlocked
#define putchar putchar_unlocked
#endif

using std::pair;
using std::vector;
using std::string;

typedef long long LL;
typedef pair<int, int> Pii;

const int oo = 0x3f3f3f3f;

template<typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, true : false; }
template<typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, true : false; }
string procStatus()
{
    std::ifstream t("/proc/self/status");
    return string(std::istreambuf_iterator<char>(t), std::istreambuf_iterator<char>());
}
template<typename T> T read(T &x)
{
    int f = 1;
    char ch = getchar();
    for (; !isdigit(ch); ch = getchar())
        f = (ch == '-' ? -1 : 1);
    for (x = 0; isdigit(ch); ch = getchar())
        x = 10 * x + ch - '0';
    return x *= f;
}
template<typename T> void write(T x)
{
    if (x == 0) {
        putchar('0');
        return;
    }
    if (x < 0) {
        putchar('-');
        x = -x;
    }
    static char s[20];
    int top = 0;
    for (; x; x /= 10)
        s[++top] = x % 10 + '0';
    while (top)
        putchar(s[top--]);
}
// EOT

const int MAXN = 1e5 + 5;

int N;
int A[MAXN];

void input()
{
    read(N);
    for (int i = 1; i <= N; ++i) {
        read(A[i]);
    }
}

void solve()
{
    LL sum = 0;
    for (int i = 1; i <= N; ++i) {
        sum += A[i];
    }
    LL atom = (LL)(1 + N) * N / 2;
    if (sum % atom != 0) {
        puts("NO");
        return;
    }
    int tot = sum / atom;

    A[0] = A[N];
    for (int i = 1; i <= N; ++i) {
        LL d = A[i] - A[i - 1];
        if (tot - d < 0 || (tot - d) % N != 0) {
            puts("NO");
            return;
        }
    }

    puts("YES");
}

int main()
{
#ifdef __DEBUG
    freopen("B.in", "r", stdin);
    freopen("B.out", "w", stdout);
#endif

    input();
    solve();

    return 0;
}

// 一郡官闲唯副使，一年冷节是清明。
//     -- 王禹偁《清明日独酌》