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>
#include <cmath>
#include <algorithm>
#include <vector>
#include <string>
#include <queue>

using namespace std;
typedef long long ll;
#define FOR(i, l, r) for(ll i = l; i <= r; i++)

int main(){
	ll n, a, b, c, d;
	cin >> n >> a >> b >> c >> d;
	string s; cin >> s;
	vector <ll> D;
	ll pos[200005];
	ll num = 0;
	ll thang = 0;
	FOR(i, 0, s.length() - 1){
		if(s[i] == '.') num++;
		else if(s[i] == '#'){
			pos[thang] = i; 
			D.push_back(num);
			num = 0;
			thang++;
		}
		if(i == s.length() - 1){
			D.push_back(num);
			num = 0;
		}
	}
	ll ko = 0;
	vector<ll> O;
	FOR(i, 0, thang){
		if(D[i] == 0){
			ko++;
			if((pos[i] < c && pos[i] > a) || (pos[i] < d && pos[i] > b)) O.push_back(pos[i]);
		}
	}
	if(ko != 0){
		if(O.size()){
			cout << "No";
			return 0;
		}	
	}
	ll helpful = 0, vtbd = 0;
	FOR(i, 0, thang){
		if(D[i] >= 3){
			helpful++;
			if((pos[i] > c || pos[i] < a) || (pos[i] < b || pos[i] > d)) vtbd++;
		}
	}
	if((a < b && b < c && c < d) || (a < c && c < b && b < d)) cout << "Yes";
	else{
		if(helpful == 0) cout << "No";
		else{
			if(vtbd) cout << "No";
			else cout << "Yes";
		}
	}
}