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 <iostream>
#include <cassert>
#include <climits>
#include <bitset>
#include <stack>
#include <queue>
#include <iomanip>
#include <limits>
#include <string>
#include <cmath>
#include <set>
#include <map>
#include <math.h>
#include <algorithm>
#include <vector>
#include <string.h>
 
using namespace std;
typedef long long ll;
 
 
typedef pair<ll,ll> P;
long long int INF = 1e18;
double Pi = 3.141592653589;
long long int mod = 1000000007;
// memset(a,0,sizeof(a)); →全部０にする
 
vector<ll> G[100005];
vector<P> tree[100010];
priority_queue <ll> pql;
priority_queue <P> pqp;
//big priority queue
priority_queue <ll,vector<ll>,greater<ll> > pqls;
priority_queue <P,vector<P>,greater<P> > pqps;
//small priority queue
//top pop

int dx[8]={1,0,-1,0,1,1,-1,-1};
int dy[8]={0,1,0,-1,1,-1,-1,1};
//↓,→,↑,←
 
#define p(x) cout<<x<<endl;
#define el cout<<endl;
#define pe(x) cout<<x<<" ";
#define ps(x) cout<<fixed<<setprecision(25)<<x<<endl;
#define pu(x) cout<<x;
#define re(i,a,b) for(i=a;i<=b;i++);
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define deba(x) cout<< #x << " = " << x <<endl

ll rui(ll abc,ll bed){
	//aのb乗を計算する
	if(bed==0){return 1;}
	else{
		ll ced = rui(abc,bed/2);
		ced *= ced;
		ced %= mod;
		if(bed%2==1){ced*=abc; ced%=mod;}
		return ced;
	}
} 

ll i,j,k,ii,jj;
ll n,m,num,sumxor,sum;
ll ans;
ll a,b,c,d,e,f,g,h,w,v;
char s[500005];
ll x[800005],y[800005],z[900005];
ll sex[500005];

int main(){	
	cin>>n;
	for(i=0;i<n;i++){
		cin>>a;
		if((a-i)%2==0){
			b++;
		}
	}
	if(b>=1){
		p("No");
	}else{
		p("Yes");
	}


	

	
	return 0;
}