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;
using ll=long long;
using ull=unsigned long long;
using pii=pair<int,int>;

#define INF LONG_MAX
#define MOD 1000000007
#define rng(a) a.begin(),a.end()
#define rrng(a) a.end(),a.begin()

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N,A,B,C,D;
  string S;
  cin>>N>>A>>B>>C>>D>>S;
  A--;
  B--;
  C--;
  D--;
  bool okb=true;
  for(int i=B;i<D;i++){
    if(S[i]=='#'&&S[i+1]=='#')okb=false;
  }
  bool oka=true;
  for(int i=A;i<C;i++){
    if(S[i]=='#'&&S[i+1]=='#')oka=false;
  }
  bool one=(oka&&okb);

  okb=true;
  for(int i=B;i<D;i++){
    if(S[i]=='#'&&S[i+1]=='#')okb=false;
  }
  S[D]='#';
  oka=true;
  for(int i=A;i<C;i++){
    if(S[i]=='#'&&S[i+1]=='#')oka=false;
  }
  bool two=(oka&&okb);

  S[D]='.';
  bool three=false;
  for(int i=B-1;i<D-2;i++){
    if(S[i]=='.'&&S[i+1]=='.'&&S[i+2]=='.')three=true;
  }
  if(C<D&&one)cout<<"Yes"<<endl;
  else if(D<C&&(three||two))cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
  return 0;
}