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<string.h>
#include<stdlib.h>
#include<math.h>
long long cmpfunc (const void * a, const void * b) {
   return ( *(long long*)a - *(long long*)b );
}



int main(void){
   
long long int test,i,j,n,count,flag=0,o1=0,o2=0,b1,x,m,l,max,sum2,min,f,c,r,o,d,sum1,sum=0,y,a,b,count1=0,a2,b2;
char k[1000000];
scanf("%lld%lld%lld%lld%lld%s",&n,&a,&b,&c,&d,k);
flag=0;
if(a>b){
    k[c-1]='#';
    for(i=b-1;i<d-2;){
        if(k[i+1]=='#'&& k[i+2]=='#'){
            printf("No");
           return 0;
        }else{
            if(k[i+1]=='#'){
                i+=2;
            }else{
                i++;
            }
        }
    }
}else{
     k[d-1]='#';
    for(i=a-1;i<c-2;i++){
        if(k[i+1]=='#'&& k[i+2]=='#'){
            printf("No");
          return 0;
        } else{ if(k[i+1]=='#'){
                i+=2;
            }else{
                i++;
          }  }
    }
}
printf("Yes");
return 0;
}