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;

typedef long long ll;
typedef vector<int> vi;
typedef pair<int,int> pi;

#define F first
#define S second
#define PB push_back
#define MK make_pair
#define REP(i, a, b) for (int i=a; i < b; i++)


bool sort_by(pi c, pi d){
	return c.S < d.S;
}

int main() {
	int n, a, b, c, d;
	string s;
	cin >> n >> a >> b >> c >> d >> s;
	string ans = "Yes";
	char p;
	REP(i, a, c+1){
		if(i != a && p == '#' && s[i] == '#'){
			cout << "No";
			exit(0);
		}
		p = s[i];
	}
	REP(i, b, d+1){
		if(i != b && p == '#' && s[i] == '#'){
			cout << "No";
			exit(0);
		}
		p = s[i];
	}
	ans = "No";
	if( d < c){
		for(int i = b; i <= d; i++){
			if(s[i-1] =='.' && s[i] =='.' && s[i+1] =='.') ans = "Yes";
		}
	}
	cout << ans;
		
	

}