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<algorithm>
#include<iostream>
#include<cstdio>

using namespace std;

int a[300001],n;

bool check(int o,int p)
{
	int mx=0,mm=p+1,s=0,ss=0;
	for(int i=o;i<=p;i++) mx=max(mx,a[i]),mm=min(mm,a[i]);
	if((mx!=p)||(mm!=o)) return false;
	for(int i=o;i<=p;i+=2)
	    if(a[i]>s) s=a[i];else if(a[i]>ss) ss=a[i];else return false;
	return true;
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	a[n+1]=n+1;
	for(int i=1;i<=n;i++)
	{
		if(a[i]==i) continue;
		int p;
		for(p=i;p<=n;p+=2)
		    if((a[p+1]!=(p+1))||(a[p+2]==(p+2))) break;
		if(!check(i,p)) {cout<<"No"<<endl;return 0;}
		i=p;
	}
	cout<<"Yes"<<endl;
	return 0;
}