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>
#include<string.h>
int main()
{
	int N;
	int A, B, C, D;
	char(*S)= (char(*))malloc(200010 * sizeof(char));
	int i;
	int c_flag = 0;
	int d_flag = 0;
	int cd_flag = 0;
	int none_count = 0;
	scanf("%d%d%d%d%d", &N, &A, &B, &C, &D);
	S[0] = '1';/*ダミー*/
	for(i = 0;i <= N+1;i++)
	{
		scanf("%c", &S[i]);/*S[0]は値なし。S[1]から*/
	}
	
	
	if (C < D)
	{
		for (i = B+1;i < D;i++)
		{
			if (S[i] == '#'&&S[i + 1] == '#')
			{
				d_flag = 1;
				break;
			}
		}
		for (i = (A + 1);i < C;i++)
		{
			if ((S[i] == '#'&&S[i + 1] == '#')||(i==D && S[i + 1] == '#')||(S[i] == '#'&& (i + 1) == D))
			{
				c_flag = 1;
				break;
			}
		}
	}else
	{
		for (i = (B + 1);i < D;i+=3)
		{
			if (S[i] == '.'&&S[i + 1] == '.'&&S[i + 2] == '.')
			{
				none_count = 1;
				break;
			}
		}
		for (i = (B + 1);i < D;i++)
		{
			if (S[i] == '#'&&S[i + 1] == '#')
			{
				d_flag = 1;
				break;
			}
		}
		for (i = (A + 1);i < C;i++)
		{
			
			if (S[i] == '#'&&S[i + 1] == '#')
			{
				c_flag = 1;
				break;
			}
			if ( (i == D && S[i + 1] == '#') || (S[i] == '#'&& (i + 1) == D))
			{
				c_flag = 1;
				cd_flag = 1;
				break;
			}
		}
	}

	if (none_count == 1 && cd_flag == 1)c_flag = 0;
	
	if (c_flag == 1 || d_flag == 1)printf("No\n");
	else printf("Yes\n");
	return 0;
}