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: Runtime Error)

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();
    int b = sc.nextInt();
    int c = sc.nextInt();
    int d = sc.nextInt();
    int[] mas = new int[n];
    String s = sc.next();
    for(int i = 0; i < n; i++) {
      if(s.charAt(i) == '#') mas[i] = 1;
    }
    String ans = "Yes";
    int p = 0;
    for(int i = a + 1; i <= c; i++) {
      if((mas[i - 1] == 1) && (mas[i] == 1)) p++;
    }
    for(int i = b + 1; i <= d; i++) {
      if((mas[i - 1] == 1) && (mas[i] == 1)) p++;
    }
    if(c > d) {
      if(p == 0) {
        int q = 0;
        for(int i = b; i <= d; i++) {
          if((mas[i - 1] == 0) && (mas[i] == 0) && (mas[i + 1] == 0)) q++;
        }
        if(q == 0) ans = "No";
      } else {
        ans = "No";
      }
    } else {
      if(p > 0) ans = "No";
    }
    System.out.println(ans);
  }
}