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;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
#define FOR(NAME, START, NUM) for (ll NAME = START; NAME < (NUM); NAME++)
#define REP(NAME, NUM) for (ll NAME = 0; NAME < (NUM); NAME++)
#define BREP(NAME, NUM) for (ll NAME = (NUM)-1; NAME >= 0; NAME--)
#define ALL(NAME) (NAME).begin(), (NAME).end()
const ll INF = 1000000000000000;

int n,a,b,c,d;
int main(){
    string s;
	cin >> n >> a >> b >> c >> d >> s;
	FOR(i,a-1,c-1){
		if(s.at(i) == '#' && s.at(i+1) == '#'){
			cout << "No" << endl;
			return 0;
		}
	}
	FOR(i,b-1,d-1){
		if(s.at(i) == '#' && s.at(i+1) == '#'){
			cout << "No" << endl;
			return 0;
		}
	}
	if(c > d){
		FOR(i,0,n-2){
			if(s.at(i) == '.' && s.at(i+1) == '.' && s.at(i+2) == '.'){
				cout << "Yes" << endl;
				return 0;
			}
		}
	}
	else{
		cout << "Yes" << endl;
		return 0;
	}

	cout << "No" << endl;

}