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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
#include <math.h>
#include <iomanip>
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
const int INF=1001001001;
vector<pair<int64_t,int64_t>>pppp(int64_t x){
  vector<pair<int64_t,int64_t>>p;
  for(int64_t i=2;i*i<=x;i++){
    int cnt=0;
    if(x%i==0){
    while(x%i==0){cnt++;x/=i;}
    p.push_back(make_pair(i,cnt));
    }
  }
  if(x!=1){p.push_back(make_pair(x,1));}
  return p; 
}
 
int main() {
  int N,A,B,C,D;
  string s;
  cin>>N>>A>>B>>C>>D>>s;
  bool ok=true;int cnt=0;
  for(int i=0;i<N;i++){
    if(s[i]!='#'){cnt++;}
    else{cnt=0;}
    if(i>=B&&cnt>=3){ok=false;}
  }
  if(C>D&&ok){cout<<"No"<<endl;}
  else{cout<<"Yes"<<endl;}
  return 0;
}