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: Time Limit Exceeded)


import java.util.Scanner;

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();
		String S = sc.next();
		S += ".";
		boolean flag = false;
		while(B <= D) {
			int Apre = A;
			int Bpre = B;
			while(Apre < C) {
				if(S.charAt(Apre) != '#' && Apre + 1 != Bpre) {
					Apre++;
				} else if(S.charAt(Apre + 1) != '#' && Apre + 2 != Bpre) {
					Apre = Apre + 2;
				} else {
					break;
				}
			}
			if(Apre == C) {
				while(Bpre < D) {
					if(S.charAt(Bpre) != '#' && Bpre + 1 != Apre) {
						Bpre++;
					} else if(S.charAt(Bpre + 1) != '#' && Bpre + 2 != Apre) {
						Bpre = Bpre + 2;
					} else {
						break;
					}
				}
			}
			if(Apre == C && Bpre == D) {
				flag = true;
				break;
			} else {
				if(S.charAt(B) != '#') {
					B++;
				} else if(S.charAt(B + 1) != '#') {
					B = B + 2;
				} else {
					B = D + 1;
				}
			}
		}
		if(flag) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}

	}

}