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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 300000;
int TLE = 300000000;
int n;
int a[MAXN+50];
void Swap(int i)
{
	if(a[i-2] > a[i-1] && a[i-1] > a[i])
		swap(a[i-2], a[i]);
}
bool Check()
{
	bool p = true;
	for(int i=1; i<=n; i++)
		p = p && a[i] == i;
	return p;
}
int main()
{
	clock_t st=clock();
	cin>>n;
	for(int i=1; i<=n; i++)
	{
		cin>>a[i];
		if((a[i]&1) != (i&1))
		{
			puts("No");
			return 0;
		}
	}
	while(TLE--)
	{	
		for(int i=n; i>=3; i--)
			Swap(i);
	}
	if(Check())
		puts("Yes");
	else
		puts("No");
}