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 = sc.nextInt() + 2;
        int A = sc.nextInt() - 1;
        int B = sc.nextInt() - 1;
        int C = sc.nextInt() - 1;
        int D = sc.nextInt() - 1;

        String str = sc.next();
        char c[] = str.toCharArray();

        //c[A] = 'A';
        //c[B] = 'B';
        //c[C] = 'C';
        //c[D] = 'D';
        boolean flag = true;
        for (int i = B; i < c.length; i++) {
            if (i + 1 == D || i + 2 == D) {
                break;
            }
            if (c[i + 1] != '.' && c[i + 2] != '.') {
                flag = false;
                break;
            }
        }
        for (int i = A; i < c.length; i++) {
            if (i + 1 == C || i + 2 == C) {
                break;
            }
            if (c[i + 1] != '.' && c[i + 2] != '.') {
                flag = false;
                break;
            }
        }

        if (C > D) {
            int num = str.indexOf("...");
            if (num == -1) {
                if (c[D - 1] == '#' || c[D + 1] == '#') {
                    flag = false;
                }
            } else if (num == D || num + 2 == D) {
                flag = false;
            } else if(num == C || num + 1 == C || num + 2 == C){
                flag = false;
            }
        }

        if (flag) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}