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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<functional>
#include<string>
#include<iomanip>
#include<utility>
#include<string>
#include<map>
#include<unordered_map>
#include<queue>

template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

using namespace std;

typedef long long ll;
typedef pair<int, int> Pi;
int gcd(int a, int b){return b ? gcd(b, a%b) : a;}
int lcm(int a, int b){return a*b / gcd(a, b);}


int main(){
    int n, a, b, c, d;
    cin >> n >> a >> b >> c >> d;
    string s;
    cin >> s;

    a--, b--, c--, d--;
    bool throuth = true;
    if(c < d){
        for(int i = a; i <= c-1; i++){
            if(s[i] == '#' && s[i+1] == '#'){
                throuth = false;
            }
        }
        for(int i = b; i <= d-1; i++){
            if(s[i] == '#' && s[i+1] == '#'){
                throuth = false;
            }
        }
    }
    else{
        for(int i = b+1; i <= d-1; i++){
            if(s[i-1] != '#' && s[i+1] != '#'){
                throuth = true;
                break;
            }
            throuth = false;
        }
    }

    if(throuth == true){
        cout << "Yes" << endl;
    }
    else{
        cout << "No" << endl;
    }
    return 0;
}