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 Java (OpenJDK 11.0.6) (judge verdict: Wrong Answer)

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