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: Accepted)

#include<bits/stdc++.h>
using namespace std;
int a[20000005];
int main()
{
	int n,A,b,c,d;
	cin>>n>>A>>b>>c>>d;
	char x;
	for(int i=1;i<=n;i++)
	{
		cin>>x;
		if(x=='.')
			a[i]=0;
		if(x=='#')
			a[i]=1;
	}
	for(int i=min(A,b);i<=max(c,d)-1;i++)
	{
		if(a[i]==a[i+1]&&a[i]==1)
		{
			cout<<"No"<<endl;
			return 0;
		}
	}
	if(A>b)
	{
		swap(A,b);
		swap(c,d);
	}
	int i=A,j=b,flag=0;
	while(i<c)
	{
		if(a[i+2]==1)
			i+=1;
		else
			i+=2;
		if(i==j)
		{
			if(j==d&&i==c)
			{
				cout<<"Yes"<<endl;
				return 0;
			}
			if(j==d)
			{
				cout<<"No"<<endl;
				return 0;
			}
			else
			{
				if(a[j+1]!=1)
					j+=1;
				else
					j+=2;
			}
		}
//		cout<<i<<" "<<j<<endl;
	}
	for(int k=j;k<=d;k++)
	{
		if((a[k+1]==1&&k+2==c)||(a[k+2]==1&&k+1==c))
		{
			cout<<"No"<<endl;
			return 0;
		}
	}
	cout<<"Yes"<<endl;
	return 0;
}