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)

/*author: hyperion_1724
  date: 
*/
#include <bits/stdc++.h>
using namespace std;
#define ll long long 
#define READ(X) cin>>X;
#define READV(X) long long X; cin>>X;
#define READAR(A,N) long long A[N]; for(int i=0;i<N;i++) {cin>>A[i];}
#define rz(A,N) A.resize(N);
#define sz(X) (long long)(X.size())
#define pii pair<int,int>
#define pll pair<int,int>
#define fi first
#define se second
#define trav(x,A) for(auto &x:A)
#define veci vector<int>
#define vecl vector<long long>
#define FORI(a,b,c) for(int a=b;a<c;a++)
#define FORD(a,b,c) for(int a=b;a>c;a--)
#define adj(N) vector<vector<int>> ADJ(N+1)


int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	

	READV(N);
	READV(A);
	READV(B);
	READV(C);
	READV(D);
	string S;
	READ(S);
	if(D>C)
	{
		bool a=true;
		FORI(i,B-1,D-1)
		{
			if(S[i]=='#' && S[i+1]=='#')
			{
				a=false;
			}
		}
		FORI(i,A-1,C-1)
		{
			if(S[i]=='#' && S[i+1]=='#')
			{
				a=false;
			}
		}
		if(a)
		{
			cout<<"YES"<<endl;
		}
		else
		{
			cout<<"NO"<<endl;
		}
	}
	else
	{
		bool a=true;
		FORI(i,A-1,C-1)
		{
			if(S[i]=='#' && S[i+1]=='#')
			{
				a=false;
			}
		}
		bool b=false;
		FORI(i,max(B-2,(ll)0),min(D-1,N-2))
		{
			if(S[i]=='.' && S[i+1]=='.' && S[i+2]=='.')
			{
				b=true;
			}
		}
		if(a&&b)
		{
			cout<<"YES"<<endl;
		}
		else
		{
			cout<<"NO"<<endl;
		}
	}

	return 0;
}