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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

import java.util.*

fun main(args: Array<String>) {
    val scanner = Scanner(System.`in`)
    val N = scanner.nextLong()
    var sum: Long = 0L
    val AN: List<Long> = (0..N - 1).map { val l = scanner.nextLong();sum += l;return@map l }
    val once = (N * (N + 1)) / 2
    if (sum % once != 0L) {
        println("NO")
        return
    }
    val k: Long = sum / once
    var count: Long = 0
    (0..N - 1).forEach { index ->
        var di = AN[((index + 1) % N).toInt()] - AN[index.toInt()] - k
        while (di < 0) {
            di += N
            count += 1
        }
        if (di != 0L) {
            println("NO")
            return
        }
    }
    if (k == count) println("YES")
    else {
        println("NO")
    }
}