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: Wrong Answer)

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

#define ll long long
#define vi vector<int>
#define vl vector<ll>
#define pii pair<int,int>
#define pll pair<ll,ll>
// pll pp; -> (pp.first,pp.second)
// vector<vl> pp(n,vl(2)); sort(pp.begin(),pp.end(),[](vl a,vl b){return a[0]<b[0];});

#define REP(i,n) for(ll (i)=0;(i)<(n);(i)++)
#define RREP(i,n) for(ll (i)=(n)-1;(i)>=0;(i)--)
#define FOR(i,a,b) for(ll (i)=(a);(i)<(b);(i)++)
#define FORR(i,a,b) for(ll (i)=(b)-1;(i)>=(a);(i)--)
#define DEBUG(x) cout << #x << ": " << x << endl
#define CHMAX(a,b) a=max((a),(b))
#define CHMIN(a,b) a=min((a),(b))
/*}}}*/
int main() {
  ll N,c,l,r,tp,d;
  vl p;
  cin >> N;
  REP(i,N){
		int x;
		cin >> x;
		p.push_back(x);
	}
  c = 0;
  REP(i,N-1){
    if(p[i]>=p[i+1]){
      c = c+1;
      if(c==1){
        l = i;
      }
      if(c==2){
        r=i;
      }
    }
  }
  if(c>2){
    cout << "NO" << endl;
  }else{
    if(c==1){
      tp = p[l];
      p[l] = p[l+1];
      p[l+1] = tp;
      d=0;
        REP(i,N){
          if(p[i]>=p[i+1]){
            d = d+1;
          }
        }
        if(d==0){
          cout << "YES" << endl;
        }else{
          cout << "NO" << endl;
        }
    }else{
      if(c==0){
        cout << "YES" << endl;
      }else{
        tp = p[l];
        p[l] = p[r+1];
        p[r+1] = tp;
        d=0;
        REP(i,N){
          if(p[i]>=p[i+1]){
            d = d+1;
          }
        }
        if(d==0){
          cout << "YES" << endl;
        }else{
          cout << "NO" << endl;
        }
      }
    }
  }
  return 0;
}