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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = Integer.parseInt(sc.next());
		int a = Integer.parseInt(sc.next())-1;
		int b = Integer.parseInt(sc.next())-1;
		int c = Integer.parseInt(sc.next())-1;
		int d = Integer.parseInt(sc.next())-1;
		
		char []p = sc.next().toCharArray();
		
		int len = 0;
		int maxl = 0;
		boolean path = true;
		
		for(int i = 0 ; i < n ;i++) {
			if(a > i || Math.max(c, d) < i) {
				continue;
			}
			
			if(p[i] == '#') {
				if(!path) {
					System.out.println("No");
					System.out.println('#');
					return;
				}
				maxl = Math.max(maxl, len);
				len = 0;
				path = false;
			}
			else {
				if(i >= b) {
				len++;
				}
				path = true;
			}
		}
		maxl = Math.max(maxl,len);
		//System.out.println(maxl);
		
		if(maxl >= 3) {
			System.out.println("Yes");
		}
		else if(c < d) {
			System.out.println("Yes");
		}
		else {
			System.out.println("No");
		}
	}

}