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<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<ctime>
#include<utility>
#include<functional>
#include<cmath>
#include<vector>
//using namespace std;
using std::min;
using std::max;
using std::swap;
using std::sort;
using std::reverse;
using std::random_shuffle;
using std::lower_bound;
using std::upper_bound;
using std::unique;
using std::vector;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef std::pair<int,int> pii;
typedef std::pair<ll,ll> pll;
void open(const char *s){
#ifndef ONLINE_JUDGE
	char str[100];sprintf(str,"%s.in",s);freopen(str,"r",stdin);sprintf(str,"%s.out",s);freopen(str,"w",stdout);
#endif
}
int rd(){int s=0,c,b=0;while(((c=getchar())<'0'||c>'9')&&c!='-');if(c=='-'){c=getchar();b=1;}do{s=s*10+c-'0';}while((c=getchar())>='0'&&c<='9');return b?-s:s;}
void put(int x){if(!x){putchar('0');return;}static int c[20];int t=0;while(x){c[++t]=x%10;x/=10;}while(t)putchar(c[t--]+'0');}
int upmin(int &a,int b){if(b<a){a=b;return 1;}return 0;}
int upmax(int &a,int b){if(b>a){a=b;return 1;}return 0;}
const int N=300010;
int n;
int a[N],b[N];
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&a[i]);
	int sum=0;
	for(int i=1;i<=n;i++)
		if(a[i]==i)
		{
			b[i]=1;
			sum++;
		}
	if(sum==n)
	{
		printf("Yes\n");
		return 0;
	}
	int l=1,r=n;
	while(b[l])
		l++;
	while(b[r])
		r--;
	for(int i=l;i<=r;)
	{
		if(i==r||!b[i+1])
		{
			printf("No\n");
			return 0;
		}
		int j=i+1;
		for(;j<=r&&b[j];j+=2)
		{
			if(b[j+1])
			{
				printf("No\n");
				return 0;
			}
		}
		j-=2;
		int s1=0,s2=0;
		for(int k=i;k<=j+1;k+=2)
			if(a[k]>k)
			{
				if(a[k]<s1)
				{
					printf("No\n");
					return 0;
				}
				s1=a[k];
			}
			else
			{
				if(a[k]<s2)
				{
					printf("No\n");
					return 0;
				}
				s2=a[k];
			}
		i=j+2;
	}
	printf("Yes\n");
	return 0;
}