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: Accepted)

#include <bits/stdc++.h>
#include <iomanip>
using namespace std;

int main(){
  int N,A,B,C,D;
  cin >> N >> A >> B >> C >> D;
  string S;
  cin >> S;
  bool ans=1;
  if(B<C&&C<D){
    for(int i=A-1;i<D-1;i++){
      if(S[i]=='#'&&S[i+1]=='#')
        ans=0;
    }
    //Aマス目からDマス目まで2連の岩があればNo
    if(ans)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
  }
  else if(B<D&&D<C){
    ans=0;
    for(int j=B-2;j<D-1;j++){
      if(S[j]=='.'&&S[j+1]=='.'&&S[j+2]=='.')
        ans=1;
    }
    //Bマス目からDマス目まで3連の空きがなければNo
    for(int k=A-1;k<C-1;k++){
      if(S[k]=='#'&&S[k+1]=='#')
        ans=0;
    }
  if(ans)
    cout << "Yes" << endl;
  else
    cout << "No" << endl;
  }
  else{
    for(int l=A-1;l<C-1;l++){
      if(S[l]=='#'&&S[l+1]=='#')
        ans=0;
    }
    for(int m=B-1;m<D-1;m++){
      if(S[m]=='#'&&S[m+1]=='#')
        ans=0;
    }
    if(ans)
      cout << "Yes" << endl;
    else
      cout << "No" << endl;
  }
}