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>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define P pair<int,int>
#define inf 1000000007
//P pa(114,514);   ,vector<P>vec(N,0);
//vector<vector<int>> vec(縦, vector<int>(横));
#define REP(i,N) for(int i=0;i<N;i++)
int gcd(int a, int b) {
    if(a>b){
        swap(a,b);
        }
    while(b%a!=0){
        b%=a;
        swap(a,b);
    }
    return a;
}
//cout << setprecision(123);
int main(){
  int N,A,B,C,D;
  string S;
  bool ok=false,dam=true,damdam=false;
  cin>>N>>A>>B>>C>>D>>S;
  
  for(int i=A;i<max(C,D)-3;i++){
    if(S[i]=='#' && S[i+1]=='#'){
      dam=false;
      break;
    }
    if(i>B-3 && i<D-1 && S[i]=='.' && S[i+1]=='.' && S[i+2]=='.') damdam=true;
  }
    
    
  if(dam){
    if(C<D){
      ok=true;
    }
    /*else if(S[B-2]=='.' && S[B]=='.'){
      ok=true;
  }*/
    else if(damdam) ok=true;
  }
  
  if(ok)cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
}