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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(){
    int n,a,b,c,d,l=0;
    int yn=0;
    char s[200000];
    scanf("%d %d %d %d %d %d",&n,&a,&b,&c,&d);
    n--;
    scanf("%c",s);
    s[a],s[b],s[c],s[d]='.';
    for(l=a;l<=n;l=l+2){
        //sunuke
        if(s[l]=='#'){
            if(s[l-1]=='#'){
                break;
            }else{
                l--;
            }
        }
        if(l==c){
            yn=1;
            break;
        }
    }
    if(yn==1){
        for(l=b;l<=n;l=l+2){
        //hunuke
            if(s[l]=='#'){
                if(s[l-1]=='#'){
                    break;
                }else{
                    l--;
                }
            }
            if(l==d){
                yn=1;
                break;
            }
        }
    }
    if(yn==1){
        printf("Yes");
    }else{
        printf("No");
    }
    return 0;
}