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 OCaml (4.02.3) (judge verdict: Time Limit Exceeded)

open Printf open Scanf
let ident v = v
let max_listi ls = 
	let rec f0 i l v vi = match l with
	| [] -> (v,vi)
	| h::t ->
		if (h > v)
			then f0 (i+1) t h i
			else f0 (i+1) t v vi
	in f0 0 ls min_int (-1)
let () =
	let n = scanf " %d" ident
	in let ar = Array.init n (fun _ -> Scanf.scanf " %d" (fun v -> v))
	in
	let f = ref true
	in
	while !f do
		let (vmax,imax) = max_listi (Array.to_list ar)
		in
		(* printf "%d,%d\n" vmax imax; *)
		for i=0 to n-1 do
			let r =
				if i=imax then ar.(i) - n
				else ar.(i) - ((i - imax + 2*n) mod n)
			in ar.(i) <- r;
			if r<=0 then f := false;
		done;
	done;
	Array.fold_left (fun u v -> (
		(* printf "%d," v; *)
		(u && v=0))) true ar
	|> (fun v -> if v then "YES" else "NO")
	|> printf "%s\n"