Problem:
Score : 1200 points

Problem Statement

You are given a permutation of 1,2,...,N: p_1,p_2,...,p_N. Determine if the state where p_i=i for every i can be reached by performing the following operation any number of times:

Choose three elements p_{i-1},p_{i},p_{i+1} (2\leq i\leq N-1) such that p_{i-1}>p_{i}>p_{i+1} and reverse the order of these three.

Constraints

3 \leq N \leq 3 × 10^5

p_1,p_2,...,p_N is a permutation of 1,2,...,N.

Input

Input is given from Standard Input in the following format:

N
p_1
:
p_N

Output

If the state where p_i=i for every i can be reached by performing the operation, print Yes; otherwise, print No.

Sample Input 1

5
5
2
1
4
3

Sample Output 1

Yes

The state where p_i=i for every i can be reached as follows:

Reverse the order of p_1,p_2,p_3. The sequence p becomes 1,2,5,4,3.

Reverse the order of p_3,p_4,p_5. The sequence p becomes 1,2,3,4,5.

Sample Input 2

4
3
2
4
1

Sample Output 2

No

Sample Input 3

7
3
2
1
6
5
4
7

Sample Output 3

Yes

Sample Input 4

6
5
3
4
1
2
6

Sample Output 4

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define N 300003

int n,k;
int a[N],q[100*N];

bool check(){
	for (int i=1;i<=n;i++)
	  if (a[i]!=i) return false;
	return true;	
}

int main(){
    scanf("%d",&n);
    for (int i=1;i<=n;i++)
      scanf("%d",&a[i]);
    int head=0,tail=0;
    for (int i=1;i<=n-2;i++)
      if (a[i]>a[i+1] && a[i+1]>a[i+2]) q[++tail]=i;
    int k=0;
    while (head<tail){
    	k++;
    	if (k>=20000000){
    		printf("Yes\n");
			return 0;	
    	}
    	int u=q[++head];
		swap(a[u],a[u+2]);
		for (int i=max(u-3,1);i<=min(u+3,n-2);i++)
		  if (a[i]>a[i+1] && a[i+1]>a[i+2]) q[++tail]=i;
    }
    if (check()==true) printf("Yes\n");
      else printf("No\n");
    return 0;
}