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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

void swap(int* a,int* b){
    int tmp=*a;
    *a=*b;
    *b=tmp;
}
int main() {
    int N,p[300001];
    scanf("%d",&N);
    for (int i = 0; i < N; ++i) {
        scanf("%d",&p[i]);
    }
    for (int i = 0; i < N; ++i) {
        if(i%2 == p[i]%2){
            printf("%s","No");
            return 0;
        }
    }
    for (int i = 0; i < N-2; ++i) {
        int flg=0;
        for(int j=i;j<N-2;++j){
            if(p[i]>p[i+1] && p[i+1]>p[i+2]){
                swap(&p[i],&p[i+2]);
                i++;
                flg=1;
            }
        }
        if(flg==0)break;
    }
    for (int i = 0; i < N; ++i) {
        if(i+1!=p[i]){
            printf("%s","No");
            return 0;
        }
    }
    printf("%s","Yes");
}