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.Arrays;
import java.util.Scanner;

public class Main {
public static void main(String[] args) {
		
		Scanner keyboard = new Scanner(System.in); 
		
		int N  = keyboard.nextInt();
		int A  = keyboard.nextInt();
		int B  = keyboard.nextInt();
		int C  = keyboard.nextInt();
		int D  = keyboard.nextInt();
		String S = keyboard.next();
		
		String[] masu = S.split("");
		
		String ans = "Yes";
		if(C > D) {
			if(masu[D-2].equals("#")) {
				ans = "No";
			}else if(masu[D-2].equals("#")&& masu[D].equals("#")) {
				ans = "No";
			}
		}
		
		int count = 1;
		for(int m = B-1; m < N-1; m++) {
			if(masu[m+1].equals("#")) {
				count = 0;
			} else {
				count++;
				if(count == 3) {
					ans = "Yes";
					break;
				}
			}
		}
		
		int StartB = B;	

		if(ans.equals("Yes")) {
			for(int i = B-1; i < D-1; i++) {
				if(masu[i+1].equals(".")) {
					StartB++;
					if(StartB == D) {
						ans = "Yes";
						break;
					}else if(StartB > D) {
						ans = "No";
						break;
					}
				}else if(masu[i+2].equals(".")) {
					StartB+=2;
					if(StartB == D) {
						ans = "Yes";
						break;
					}else if(StartB > D) {
						ans = "No";
						break;
					}
				}else {
					ans = "No";
					break;
				}
			}
		}
	
		int StartA = A;
		
		if(ans.equals("Yes")) {
			if(StartA != 1) {
				for(int j = A-1; j < D-1; j++) {
					if(masu[j+1].equals(".")) {
						StartA++;
						if(StartA == C) {
							ans = "Yes";
							break;
						}else if(StartA > C) {
							ans = "No";
							break;
						}
					}else if(masu[j+2].equals(".")) {
						StartA+=2;
						if(StartA == C) {
							ans = "Yes";
							break;
						}else if(StartA > C) {
							ans = "No";
							break;
						}
					}else {
						ans = "No";
						break;
					}
				}
			}else {
				for(int k = A; k < D-1; k++) {
					if(masu[k+1].equals(".")) {
						StartA++;
						if(StartA == C) {
							ans = "Yes";
							break;
						}else if(StartA > C) {
							ans = "No";
							break;
						}
					}else if(masu[k+2].equals(".")) {
						StartA+=2;
						if(StartA == C) {
							ans = "Yes";
							break;
						}else if(StartA > C) {
							ans = "No";
							break;
						}
					}else {
						ans = "No";
						break;
					}
				}
			}
		}
		
		System.out.println(ans);
		keyboard.close();	
	}
}