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<stdlib.h>

int main()
{
	int N,A,B,C,D;
	char S[200005];
	int i,j;
	scanf("%d %d %d %d %d",&N,&A,&B,&C,&D);
	for(i=0;i<N;i++) scanf("%c",&S[i]);
	if(C<D)
	{
	// Move B first to D
	  for(i=B;i<D;i++) 
	  {
		  if((S[i-1]=='#')&(S[i]=='#'))
		  {
			  printf("No\n");
			  return 0;
		  }
	  }
	// Move A to C
	  for(i=A;i<C;i++)  
          {
                  if((S[i-1]=='#')&(S[i]=='#'))
                  {
                          printf("No\n");
                          return 0;
                  }
          }
	  printf("Yes\n");
	  return 0;
	}else{
		// Find a spot where A can pass B
	        for(i=B;i<=D;i++) 
		{
		   if((S[i-2]=='.')&(S[i-1]=='.')&(S[i])=='.')
		   {
			   // check if can move A and B there
                           for(j=A;j<C;j++) if((S[j-1]=='#')&(S[j]=='#')) break;
			   printf("Yes\n");
			   return 0;
		   }
		}
		printf("No\n");
		return 0;
	   }
	return 0;
}