Problem:
Score : 200 points

Problem Statement

We have a sequence p = {p_1,\ p_2,\ ...,\ p_N} which is a permutation of {1,\ 2,\ ...,\ N}.

You can perform the following operation at most once: choose integers i and j (1 \leq i < j \leq N), and swap p_i and p_j. Note that you can also choose not to perform it.

Print YES if you can sort p in ascending order in this way, and NO otherwise.

Constraints

All values in input are integers.

2 \leq N \leq 50

p is a permutation of {1,\ 2,\ ...,\ N}.

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

Print YES if you can sort p in ascending order in the way stated in the problem statement, and NO otherwise.

Sample Input 1

5
5 2 3 4 1

Sample Output 1

YES

You can sort p in ascending order by swapping p_1 and p_5.

Sample Input 2

5
2 4 3 5 1

Sample Output 2

NO

In this case, swapping any two elements does not sort p in ascending order.

Sample Input 3

7
1 2 3 4 5 6 7

Sample Output 3

YES

p is already sorted in ascending order, so no operation is needed.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>
#include <stdlib.h>

int main() {
	int N, *p;
	int P1, I1, P2, I2;
	int ans = 0;	//YESと表示

	//Nとp1からpnの読み込み
	scanf("%d", &N);

	p = calloc(N, sizeof(int));
	scanf("%d", &p[0]);

	for (int i = 1; i < N; i++) {
		scanf("%d", &p[i]);

		//順番の入れ替わり一つ目があったとき
		if (p[i - 1] > p[i]) {
			P1 = p[i - 1];
			I1 = i - 1;
			ans = 1;	//NOの表示に切り替えておく。順番の入れ替わりが1つのみのとき、NOと表示されるようになる。

			for (int j = I1 + 2; j < N; j++) {
				scanf("%d", &p[j]);

				//順番の入れ替わり二つ目があったとき
				if (p[j - 1] > p[j]) {
					P2 = p[j];
					I2 = j;

					p[I1] = P2;
					p[I2] = P1;
					ans = 0;	//YESの表示に切り替えておく。

					for (int k = I1; k < N; k++) {
						if (p[k - 1] > p[k]) {
							ans = 1;
							break;
						}
					}
					break;
				}
			}
			break;
		}
	}
	ans == 0 ? printf("YES\n") : printf("NO\n");
}