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>

int main(void){

    int n,a,b,c,d;

    char str[200000];

    int i,j;//制御
    int hantei,hantei2,hantei3;

    scanf("%d %d %d %d %d",&n,&a,&b,&c,&d);
    scanf("%s",str);

    if(c<d){
        for(i=1;i<d;i++){
            if(str[i]=='#'&&str[i-1]=='#'){
                hantei=0;//hantei=0はno
                break;
            }
            else{
                hantei=1;
            }
        }
        if(hantei==0){
            printf("No");
        }
        else{
            printf("Yes");
        }

    }
    else{
        str[d-1]='#';
        for(i=2;i<d;i++){
            if(str[i-2]=='.'&&str[i-1]=='.'&&str[i]=='.'){
                hantei=1;
                break;
                //printf("koko1");
            }
            else{
                hantei=0;
               // printf("koko2");
            }
        }
        for(i=d;i<c;i++){
             if(str[i]=='#'&&str[i-1]=='#'){
                hantei3=0;//hantei=0はno
               // printf("koko3");
                break;}
                else 
                hantei3=1;
            }
        for(i=1;i<d;i++){
             if(str[i]=='#'&&str[i-1]=='#'){
                hantei2=0;//hantei=0はno
               // printf("koko3");
                break;
            }
            else{
                hantei2=1;
               // printf("koko4");
            }
        }
        if(hantei==1&&hantei2==0){
            printf("Yes");
        }
        else if(hantei==1&&hantei2==1){
            printf("Yes");
        }
        else if(hantei==0&&hantei2==1){
            printf("Yes");  
            }
            else if(hantei3==0){
                printf("No");
            }
        else{
            printf("No");
        }
        
    }
    
    



    return 0;
}