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(void)
{
	
	int N,A,B,C,D;
	scanf("%d %d %d %d %d\n",&N,&A,&B,&C,&D);
	int i;
	int S[N+1];
	char w;
	for(i=1; i < N+1; i++)
	{
		scanf("%c",&w);
		if(w=='#') S[i] = 1;
		else if(w=='.') S[i] = 2;
	}
	int ans = 0;
	if(C < D)
	{
		for(i = A; i < D; i++)
		{
			if(S[i] == 1 && S[i+1] == 1)
			{
				printf("No");
				return 0;
			}
		}
		printf("Yes");
	}
	else if(D < C)
	{
		for(i = A; i < C; i++)
		{
			if(S[i] == 1 && S[i+1] == 1)
			{
				ans = 1;
			}
		}
		for(i = B; i < D; i++)
		{
			if(S[i] == 2 && S[i+1] == 2 && S[i-1] == 2 && ans == 0) ans = 2;
		}
		if(ans == 0) ans = 1;
	}
	if(ans == 2) printf("Yes");
	else if(ans == 1) printf("No");
	//printf("C = %d, D = %d, ans = %d\n",C,D,ans);
	return 0;
}