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 <math.h>


int loop(void);


int keta;
int aa[11];



int N,i,j,a,b,c,count,x3,x5,x7,x,y,z;




int A,B,C,D; 
int An, Bn;
char str[200000];
int hantei;


int main(){

hantei=0;

scanf("%d,%d,%d,%d,%d", &N, &A, &B, &C,&D);



scanf("%s", str);

loop();



if(hantei == 1){

printf("Yes");


}else{


printf("No");

} 






//if(str[2]=='a'){printf("yes");}

//printf("%s", str);

   return(0);
} 


int loop(){


   if((str[C-1] == '#') || (str[D-1] == '#')|| (str[C] == str[D]) ){
   
     return(2);
   
   } 


   if((A == C) && (B == D)){
     
       hantei =1;
       return(1);      
       
   }




   
   
   if(2 <= C && (hantei ==0)){
   
   C = C-1;
   
   x=loop();
   
   if(x==2){ C=C+1;}
   
   
   }
   
   if(3 <= C && (hantei ==0) ){   
   
   C = C-2;
   
   x=loop();
   
   if(x==2){ C=C+2;}
   
   }
   
   if(2 <= D && (hantei ==0)){
   
   D = D-1;
   
   x=loop();
   
   if(x==2){ D=D+1;}
   
   }
   
   if(3 <= D && (hantei ==0)){
   
   D = D-2;
   
   x=loop();
   
   if(x==2){ D=D+2;}
   
   }


   
   
   
   return(0);
}