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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

//type
#define ll long long
typedef pair<int, int> P;
//定数
#define INF 1000000000000 //10^12:∞
#define MOD 1000000007    //10^9+7:合同式の法
#define MAXR 100000       //10^5:配列の最大のrange
//略記
#define PB push_back //挿入
#define MP make_pair //pairのコンストラクタ
#define F first      //pairの一つ目の要素
#define S second     //pairの二つ目の要素
#define Z class
// OTHER
// xの二乗を返す (関数テンプレート版)
template <typename T>
T square(T x) { return x * x; }
#define chmax(x, y) (x = max(x, y))
#define chmin(x, y) (x = min(x, y))
// loop
#define rep(i, n) for (int i = 0; i < (n); i++)
#define REPD(i, n) for (ll i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = a; i <= ll(b); i++)
#define FORD(i, a, b) for (ll i = a; i >= ll(b); i--)
#define FORA(i, I) for (const auto &i : I)

// vector
#define ALL(x) x.begin(), x.end()

// output
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define isYES(x) printf("%s\n", (x) ? "YES" : "NO")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define ln cout << '\n'
template <Z A>
void pr(A a)
{
  cout << a;
  ln;
}
template <Z A, Z B>
void pr(A a, B b)
{
  cout << a << ' ';
  pr(b);
}

int ans = 0;
// * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
// * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *

int main()
{
  int ans = 0;
  int n;
  cin >> n;
  vector<int> v(n);
  vector<int> a(n);
  rep(i, n) cin >> v[i];
  a = v;
  sort(v.begin(), v.end());
  rep(i, n)
  {
    if (a[i] != v[i])
      ans++;
  }
  if (ans == 2 || ans == 0)
    YES();
  else
    NO();
}