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: Accepted)

#include<bits/stdc++.h>
using namespace std;
const int N=300005;
int n,a[N],b[N];
bool check(int l,int r){
	int t1=0,t2=0;
	for (int i=l;i<=r;i++)
		if (a[i]<i){
			if (t1>a[i]||a[i]<l||a[i]>r)return 0;
			t1=a[i];
		}
		else if (a[i]>i){
			if (t2>a[i]||a[i]<l||a[i]>r)return 0;
			t2=a[i];
		}
	return 1;
}
int main(){
	scanf("%d",&n);
	for (int i=1;i<=n;i++)scanf("%d",&a[i]),b[i]=a[i]==i;
	for (int i=3;i<=n;i++)
		if (!b[i-2]&&!b[i-1]&&!b[i]){
			puts("No");
			return 0;
		}
	for (int i=1,j=1;i<=n;i=j+1)
		if (!b[i]){
			j=i;int flag=0;
			while (j<n&&(b[j+1]^flag))flag^=1,j++;
			if (!check(i,j)){
				puts("No");
				return 0;
			}
		}
		else j=i;
	puts("Yes");
	return 0;
}