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.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a = sc.nextInt()-1;
        int b = sc.nextInt()-1;
        int c = sc.nextInt()-1;
        int d = sc.nextInt()-1;
        int l1 = Math.min(a, c);
        int r1 = Math.max(a, c);
        int l2 = Math.min(b, d);
        int r2 = Math.max(b, d);
        char[] s = sc.next().toCharArray();
        boolean ans = true;
        boolean cross = (a < b && d < c) || (b < a && c < d);
        int idx = l1;
        boolean tmp1 = false;
        while(true){
            if(r1 - idx < 3){
                tmp1 = true;
                break;
            }
            if(idx+2 < n && s[idx+2] != '#') idx += 2;
            else if(idx+1 < n && s[idx+1] != '#') idx += 1;
            else {
                tmp1 = false;
                break;
            }
        }
        idx = l2;
        boolean tmp2 = false;
        while(true){
            if(r2 - idx < 3){
                tmp2 = true;
                break;
            }
            if(idx+2 < n && s[idx+2] != '#') idx += 2;
            else if(idx+1 < n && s[idx+1] != '#') idx += 1;
            else {
                tmp2 = false;
                break;
            }
        }
        if(tmp1 && tmp2){
            if(cross){
                int l = 0;
                int r = 0;
                int tmp = 0;
                int max = 0;
                for (int i = 0; i < n; i++) {
                    if(s[i] == '.'){
                        r = i;
                        tmp++;
                    }else if(1 < i - r){
                        if(2 < max && (l <= Math.max(l1, l2) && Math.min(r1, r2) <= r)){
                            ans = true;
                            break;
                        }else{
                            tmp = 0;
                            max = 0;
                            l = i+1;
                            r = i+1;
                        }
                    }else{
                        if(max < tmp) max = tmp;
                        tmp = 0;
                    }
                }
                if(2 < max && (l <= Math.min(l1, l2) && Math.max(r1, r2) <= r)){
                    ans = true;
                }else ans = false;
            }else{
                ans = true;
            }
        }else{
            ans = false;
        }
        System.out.println(ans ? "Yes":"No");
        sc.close();

    }

}