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){

        long int a,b,c,d,n;
        int judge = 0;
        char s[200000];
    
        scanf("%ld%ld%ld%ld%ld",&n,&a,&b,&c,&d);
        scanf("%s",s);

        if(s[a-1] == '.' && s[b-1] == '.' && s[c-1] == '.' && s[d-1] == '.'){//abcdについて     
                if(c>d){//どこかでaがbを跨がなければならない//bとdの間に...がないとできない
                        for(int i=b-1;i<=d-1;i++){
                                if(s[i] == s[i+1] && s[i+1] == s[i+2] && s[i] == '.'){
                                        judge = 2;
                                }   
                        }   
                }else{
                        judge = 2;
                }   
        }   



/*      
                if(d>c && s[c-2] == '.' && s[c] == '.'){//cの前後は.
                        judge = 2;
                }else if(c>d && s[d-2] == '.' && s[d] =='.'){
                        judge = 2;
                }
*/


        for(int i=a-1;i<d-1;i++){
                if(s[i] == s[i+1] && s[i] == '#'){
                        judge = 1;
                }   
        }   
    
        if(judge == 2){ 
                printf("Yes\n");
        }else{
                printf("No%d\n",judge);
        }   
return 0;
}