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.*;

public class Main{
  public static void main(String[]args){
    Scanner sc = new Scanner(System.in);
    String[] S = sc.nextLine().split(" ");
    int N = Integer.parseInt(S[0]);
    int A = Integer.parseInt(S[1])-1;
    int B = Integer.parseInt(S[2])-1;
    int C = Integer.parseInt(S[3])-1;
    int D = Integer.parseInt(S[4])-1;
    
    String T = sc.nextLine();
    char[] field = new char[N];
    
    for(int i = 0; i < N; i++){
      field[i] = T.charAt(i);
    }
    
    boolean rock2 = false;
    for(int i = A; i <= C; i++){
      if(field[i] == '#' && field[i+1] == '#'){
        rock2 = true;
      }
    }
    for(int i = B; i <= D; i++){
      if(field[i] == '#' && field[i+1] == '#'){
        rock2 = true;
      }
    }
    if(rock2){
      System.out.println("No");
      return;
    }
    
    if(C < D){
      System.out.println("Yes");
      return;
    }
    
    boolean tri_check = false;
    int num = 0;
    if(A+1 == B){
      num = A;
    }else{
      num = B;
    }
    for(int i = num; i <= D-2; i++){
      if(field[i] == '.' && field[i+1] == '.' && field[i+2] == '.'){
        tri_check = true;
        break;
      }
    }
    
    boolean near_check = false;
    if(field[D-1] == '#' || field[D+1] == '#'){
      near_check = true;
    }
    
    if(tri_check){
      System.out.println("Yes");
    }else if(near_check){
      System.out.println("No");
    }else{
      System.out.println("Yes");
    }
  }
}