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 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void){
  int n ,a,b,c,d,j=0,k1=0,k2=0;
  char s[200005];
  scanf("%d %d %d %d %d",&n,&a,&b,&c,&d);
  scanf("%s",s);
  if(n==4){
    printf("Yes");
    return 0;
  }
  for(int i=0;i<n;i++){
    if(c>d){
      if((s[i]=='.')&&(s[i+1]=='.')&&(s[i+2]=='.')&&(i<n-2)&&(i<d)&&(i<n-2)&&(i>=b-2)) k1=1;
      if((s[i]=='#')&&(s[i+1]=='#'&&(i<c-1)&&(i<n-1)&&(i>=a))) k2=1;
    }else{
      if((s[i]=='#')&&(s[i+1]=='#'&&(i<d-1)&&(i<n-1)&&(i>=a))) j=1;
    }
  }
  if(c>d){
    if(k1==0||k2==1) printf("No");
    if(k1==1&&k2==0) printf("Yes");
  }else{
    if(j==1) printf("No");
    if(j==0) printf("Yes");
  }
  return 0;
}