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;
  int t = 0;
  int p = 0;
  bool k = 0;
  bool m = 0;
  for(int i = A;i<C;i++){
    if(s.at(i) == '.'){
      t = 0;
    }
    else{
      t++;
    }
    if(t>=2){
      k = 1;
    }
  }
  for(int i = B;i<D;i++){
    if(s.at(i) == '.'){
      t = 0;
    }
    else{
      t++;
    }
    if(t>=2){
      k = 1;
    }
  }
  if(C>D){
    m = 1;
    for(int k = B;k<D;k++){
      if(s.at(k) == '.'){
        p++;
      }
      else{
        p = 0;
      }
      if(p>=3){
        m = 0;
      }
    }
  }
  if(m == 0 && k == 0){
    cout << "Yes" << endl;
  }
  else{
    cout << "No" << endl;
  }
  return 0;
}