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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.*;
import static java.lang.Math.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a = sc.nextInt();
        int b = sc.nextInt();
        int c = sc.nextInt();
        int d = sc.nextInt();
        char s[] = new char[n];
        boolean flag = true;
        s = sc.next().toCharArray();
        if(c<d){
            for(int i = a; i<=c-2; i++){
                if(s[i]=='#' && s[i+1]=='#'){
                    flag = false;
                }
            }
            if(flag){
                System.out.println("Yes");
            }else{
                System.out.println("No");
            }
            return;
        }else{
            flag = false;
            for(int i = b-2; i< d-1; i++){
                if(s[i] == '.' && s[i+1] == '.' && s[i+2] == '.'){
                    flag = true;
                }
            }
            // for(int i = a; i<= n-1; i++){
            //     if(s[i]=='#' && s[i+1]=='#'){
            //         flag = false;
            //     }
            // }
            
            if(flag){
                System.out.println("Yes");
            }else{
                System.out.println("No");
            }
        }  
    }
}