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;

int main() {
    int N, A, B, C, D;
    cin >> N >> A >> B >> C >> D;
    string S;
    cin >> S;
    S = '#' + S + '#';
    
    bool flag = true;
    
    for (int i = A; i < C+1; i++){
        if (S.at(i) == S.at(i+1) && S.at(i) == '#'){
            flag = false;
        }
    }
    
    for (int i = B; i < D+1; i++){
        if (S.at(i) == S.at(i+1) && S.at(i) == '#'){
            flag = false;
        }
    }
    
    if (C - D == 1 && S.at(C+1) == '#' && S.at(D-1) == '#'){
        flag = false;
    }
    else if (C - D == 2 && S.at(C-1) == '#' && S.at(D-1) == '#'){
        flag = false;
    }
    else if (C - D == 3 && S.at(D-1) == '#' && S.at(C+1) == '#' && (S.at(C-1) == '#' || S.at(D+1) == '#')){
        flag = false;
    }
    
    
    if (D<C && C - D > 1){
        bool flag2 = false;
        for (int i=D+1; i<C-1; i++){
            if (S.at(i) == S.at(i+1) && S.at(i) == '.'){
                flag2 = true;
            }
        }
        
        if (flag2 == false){
            flag = false;
        }
    }
    
    for (int i=0; i < N-1; i++){
        if (S.at(i) == S.at(i+1) && S.at(i+1) == S.at(i+2) && S.at(i) == '.'){
            flag = true;
        }
    }
  
    if (N > 1000000){
      flag = false;
    }
    
    if (flag == true){
        cout << "Yes" << endl;
    }
    else {
        cout << "No" << endl;
    }
    
  }