Problem:
Score : 400 points

Problem Statement

There are N squares arranged in a row, numbered 1, 2, ..., N from left to right.
You are given a string S of length N consisting of . and #. If the i-th character of S is #, Square i contains a rock; if the i-th character of S is ., Square i is empty.

In the beginning, Snuke stands on Square A, and Fnuke stands on Square B.

You can repeat the following operation any number of times:

Choose Snuke or Fnuke, and make him jump one or two squares to the right. The destination must be one of the squares, and it must not contain a rock or the other person.

You want to repeat this operation so that Snuke will stand on Square C and Fnuke will stand on Square D.

Determine whether this is possible.

Constraints

4 \leq N \leq 200\ 000

S is a string of length N consisting of . and #.

1 \leq A, B, C, D \leq N

Square A, B, C and D do not contain a rock.

A, B, C and D are all different.

A < B

A < C

B < D

Input

Input is given from Standard Input in the following format:

N A B C D
S

Output

Print Yes if the objective is achievable, and No if it is not.

Sample Input 1

7 1 3 6 7
.#..#..

Sample Output 1

Yes

The objective is achievable by, for example, moving the two persons as follows. (A and B represent Snuke and Fnuke, respectively.)

A#B.#..

A#.B#..

.#AB#..

.#A.#B.

.#.A#B.

.#.A#.B

.#..#AB

Sample Input 2

7 1 3 7 6
.#..#..

Sample Output 2

No

Sample Input 3

15 1 3 15 13
...#.#...#.#...

Sample Output 3

Yes

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll; 
#define arrlen(arr) (sizeof(arr)/sizeof(arr[0]))
#define Sort(arr) sort(arr, arr+arrlen(arr))
#define For(i,s,n) for(int i=s;i<n;i++)
	
bool check1(string s){
	For(i,0,s.size()-1){
		if(s[i]=='#'&&s[i+1]=='#') return false;
	}
	return true;
}

int main(){
	int N,A,B,C,D;
	string S;
	cin >> N >> A >> B >> C >> D >> S;
	//int left,right;
	
	string sa=S.substr(A-1,B-A+1);
	string sb=S.substr(C-1,D-C+1);
	
	if(C<D){
		if(check1(sa)&&check1(sb)) cout << "Yes";
		else cout << "No";
	}
	else{
		if(!check1(sa)) cout << "No";//A-C間がダメ
		else{//A-C間はok
			for(int i=B-2;i<=D-2;i++){
				if(S[i]=='.'&&S[i+1]=='.'&&S[i+2]=='.'){
					cout << "Yes";
					return 0;
				}
			}
			cout << "No";
			return 0;
		}
	}
}