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(){
	int n,a,b,c,d,ch=1,cr=0;
  	scanf("%d %d %d %d %d ", &n,&a,&b,&c,&d);
  	char s[n];
  	a--;b--;c--;d--;
  	fgets(s,sizeof(s),stdin);
  	s[a] = 'A';s[b] = 'B';
  	int mm;
	if(c > d){ mm = c;} else{ mm = d;}
  	for(int i=a;i<mm;i++){
        if(s[i]=='#' && s[i+1]=='#' && ch==1){
          printf("No\n");
          ch=0;
        }
    }
  	if(c>d){
		if(a<n-3){
      		for(int i=a;i<n-3;i++){ 
              	if(s[i]=='#' && s[i+3]=='#' && ch==1){
              		printf("No\n"); 
                  	ch=0;
                }
            }
        }
      	for(int i=a;i<=mm;i++){
        	if(s[i]=='.' && s[i+1]=='.' && s[i+2]=='.' && ch==1){
             	printf("Yes\n");
              	ch=0;
            }
        }
    } else{
      printf("Yes\n");
    }
  	return 0;
  
}