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: Accepted)

#include<stdio.h>

int main()
{
	int N,A,B,C,D;
	char S[200005];
	int i,j,maxDC;
	scanf("%d %d %d %d %d",&N,&A,&B,&C,&D);
	scanf("%s",&S);
	maxDC=D;
	if(C>D) maxDC=C;
        for(i=A-1;i<C-1;i++)
        {
                if((S[i]=='#')&&(S[i+1]=='#'))
                {
                        printf("No\n");
                        return 0;
                }
        }
	for(i=B-1;i<D-1;i++)
        {
                if((S[i]=='#')&&(S[i+1]=='#'))
                {
                        printf("No\n");
                        return 0;
                }
        }

	if(C<D)
	{
		printf("Yes\n");
		return 0;
	}else{
		// Find a spot where A can pass B
		for(i=B-1;i<D;i++) 
		{
			if((S[i-1]=='.')&&(S[i]=='.')&&(S[i+1]=='.'))
			{
			   printf("Yes\n");
			   return 0;
			}
		}
		printf("No\n");
		return 0;
	}
}