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(){
  long long n,a,b,c,d;
  cin >> n >> a >> b >> c >> d;
  string s;
  cin >> s;
  
  int count = 0;
  bool overtake = false;
  bool can1 = false;
  bool can2 = false;
  bool answer = false;
  
  long long bigger;
  if(c > d){
    bigger = c;
  }else{
    bigger = d;
  }
  
  for(long long i=b-2; i<bigger; i++){
    if(s.at(i) == '.'){
      count++;
      if(count == 3) {
        overtake=true;
        break;
      }
    }else{
      count=0;
    }
  }
  
  int position1=a-1;
  int position2=b-1;
  
  while(1){
    if(s.at(position1+1) == '.'){
      position1++;
    }else{
      if(s.at(position1+2) =='.'){
        position1 += 2;
      }else{
        break;
      }
    }
    
    if(position1 == c-1) {
      can1 = true;
      break;
    }
  }
  
  while(1){
    if(s.at(position2+1) == '.'){
      position2++;
    }else{
      if(s.at(position2+2) =='.'){
        position2 += 2;
      }else{
        break;
      }
    }
    
    if(position2 == d-1) {
      can2 = true;
      break;
    }
  }
  
  if(can1 == true && can2 == true){
    if(c<d){
      answer = true;
    }else if(c>d){
      if(overtake == true){
        answer = true;
      }
    }
  }
  
  if(answer == true){
    cout << "Yes" << endl;
  }else{
    cout << "No" << endl;
  }
}