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;
  string S;
  
  cin >> N >> A >> B >> C >> D;
  cin >> S;
  
  vector<int> rock;
  vector<int> jump;
  
  for(int i=0; i<N-1; i++){
    
    if(S.at(i) == '#' && S.at(i+1) == '#')
      rock.push_back(i+1);

    if(i != 0){
      if(S.at(i-1) == '.' && S.at(i) == '.' && S.at(i+1) == '.')
        jump.push_back(i+1);
    }
  
  }
  

  bool flag = false;

  for(int i=0; i<rock.size(); i++){
    
    if(rock.at(i) > A && rock.at(i) < C){
      flag = true; break;
    }
    if(rock.at(i) > B && rock.at(i) < D){
      flag = true; break;
    }
  
  }
  
  if(C > D){
    //入れ替えが必要
    flag = true;
    for(int i=0; i<jump.size(); i++){
      
      if(jump.at(i) > B && jump.at(i) <= D){
        flag = false; break;
      }
    
    }
  
  }

  
  if(flag)
    cout << "No" << endl;
  else
    cout << "Yes" << endl;

}