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;
typedef bool boolean;

const int N = 3e5 + 5;

int n;
int p[N], q[N];

void work(int x) {
	if (p[x] < p[x - 1] && p[x - 1] < p[x - 2]) {
		int a = p[x], b = p[x - 2];
		swap(p[x], p[x - 2]);
		swap(q[a], q[b]);
		return;
	}
	puts("No");
	exit(0);
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", p + i);
		q[p[i]] = i;
	}
	int fin = 0;//, t = 0;
	do {
		while (fin < n && p[fin + 1] == fin + 1)
			fin++;
		if (fin < n) {
			work(q[fin + 1]);
//			t++;
		}
	} while (fin < n);
//	cerr << t << '\n';
	puts("Yes");
	return 0;
}