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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
  "bufio"
  "os"
  "strconv"
  "strings"
  "fmt"
)

var reader = bufio.NewReaderSize(os.Stdin, 1000000)
var writer = bufio.NewWriter(os.Stdout)
func NextLine() string {
  var line, buffer []byte
  var isPrefix bool = true
  var err error
  for isPrefix {
    line, isPrefix, err = reader.ReadLine()
    if err != nil { panic(err) }
    buffer = append(buffer, line...)
  }
  return string(buffer)
}
func NextInt() int {
  var n int
  n, _ = strconv.Atoi(NextLine())
  return n
}
func NextIntVec() []int {
  L := strings.Split(NextLine(), " ")
  N := make([]int, len(L))
  for i := range N {
    N[i], _ = strconv.Atoi(L[i])
  }
  return N
}
func Write(s interface{}) {
  fmt.Fprintln(writer, s)
}
func Output() {
  _ = writer.Flush()
}

func Sum(A *[]int) int {
  s := 0
  for _, a := range (*A) {
    s += a
  }
  return s
}

func main() {
  N := NextInt()
  A := NextIntVec()
  if Sum(&A) % (N * (N + 1) / 2) != 0 {
    Write("NO")
    Output()
    return
  }
  M := Sum(&A) / (N * (N + 1) / 2)
  A = append(A, A[0])
  cnt := 0
  ans := ""
  for i := 0; i < N; i++ {
    D := A[i + 1] - A[i]
    k := (M - D)/N
    if M - N * k != D {
      ans = "NO"
      break
    }
    cnt += k
  }
  if ans == "" && cnt == M {
    ans = "YES"
  } else {
    ans = "NO"
  }
  Write(ans)
  Output()
}