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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
int n,m,i,j,t,k,s,a,b,c,d;
char st[200005];

int main()
{
	scanf("%d%d%d%d%d",&n,&a,&b,&c,&d);
	scanf("%s",st+1);
	for (i=a;i<max(c,d);++i) if (st[i]=='#'&&st[i+1]=='#')
	{
		printf("No\n");
		return 0;
	}
	if (c<d)
	{
		printf("Yes\n");
		return 0;
	}
	for (i=a-2;i<=min(c,n-2);++i)
	 if (st[i]=='.'&&st[i+1]=='.'&&st[i+2]=='.')
	{
		printf("Yes\n");
		return 0;
	}
	for (i=c+2;i<=n-2&&(st[i]!='#'||st[i+1]!='#');++i)
	 if (st[i]=='.'&&st[i+1]=='.'&&st[i+2]=='.')
	{
		printf("Yes\n");
		return 0;
	}
	for (i=a-2;i>0&&(st[i]!='#'||st[i+1]!='#');--i)
	 if (st[i]=='.'&&st[i+1]=='.'&&st[i+2]=='.')
	{
		printf("Yes\n");
		return 0;
	}
	printf("No\n");
	return 0;
}