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++14 (GCC 5.4.1) (judge verdict: Accepted)

// Daniel Grzegorzewski
// while (clock()<=69*CLOCKS_PER_SEC)
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")

#define MP make_pair
#define PB push_back
#define ST first
#define ND second

using namespace __gnu_pbds;
using namespace std;

template <typename T>
using ordered_set =
    tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

//X.find_by_order(k); - zwraca iterator na k-ty element (numeracja od zerowego)
//X.order_of_key(k); - zwraca liczbę elementów ostro mniejszych niż k

typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<PII> VII;
typedef long long LL;

void init_ios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}

const int N = 55;

int n, p[N], hm;

int main() {
  init_ios();
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> p[i];
    if (p[i] != i)
      ++hm;
  }
  if (hm <= 2)
    cout<<"YES\n";
  else
    cout<<"NO\n";
}