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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
    // データ読み込み部
    fscanf(STDIN, "%d", $N);
    $A = explode(" ", trim(fgets(STDIN)));

    // 石の数の総計
    $total = array_sum($A);
    // 1回の操作で取り出す石の数
    $one = (1 + $N) * $N / 2;
    // 取り出す回数を算出
    $round = $total / $one;
    // 切れ目の数カウント用
    $cut = 0;
    // 取り出す回数が整数でない場合はNOを出力して終了
    if (floor($round) != $round) {
        echo "NO\n";
        exit;
    }

    // 配列の最後に1番目の箱をつける
    $A[] = $A[0];
    for ($i = 0; $i < $N; $i++) {
        // 「右の箱の石の数」引く「左の箱の石の数」引く「取り出す回数」
        $diff = $A[$i+1] - $A[$i] - $round;
        if ($diff < 0) {
            // ここがマイナスになる場合には切れ目があるので切れ目の回数を計算
            $kaisuu = floor($diff / $N);
            // 切れ目の数がマイナスで増えていく
            $cut += $kaisuu;
        }
    }

    // 「切れ目の数」と「取り出す回数」が同じならYES
    if ($round == -$cut) {
        echo "YES\n";
    } else {
        echo "NO\n";
    }