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++ (Clang 10.0.0) (judge verdict: Accepted)

//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<climits>

/*
#include<boost/multiprecision/cpp_long long.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/rational.hpp>
namespace mp = boost::multiprecision;
using Real = mp::number<mp::cpp_dec_float<1024>>;
using Blong long = mp::cpp_long long;
*/

using namespace std;

#define REP(i,s,e) for((i)=(s);(i)<(e);(i)++)
#define RREP(i,s,e) for((i)=((s)-1);(i)>=(e);(i)--)
#define FOR(i,n) for((i)=(0);(i)<(n);(i)++)
#define RFOR(i,n) for((i)=((n)-1);(i)>=(0);(i)--)
#define MOD 1000000007

int main(){
    int N,A,B,C,D;
    cin>>N>>A>>B>>C>>D;
    A--;
    B--;
    C--;
    D--;
    string S;
    cin>>S;
    S=S+"###";

    bool f=false;

    for(int i=B;i<D;i++){
        if(S[i-1]=='.'&&S[i+1]=='.'&&S[i]=='.') f=true;
        if(S[i]=='.'&&S[i+1]=='.'&&S[i+2]=='.') f=true;
        if(i+1==D||i+2==D) break;
        if(S[i+1]=='.');
        else if(S[i+2]=='.'){
            i++;
        }
        else {
            cout<<"No"<<endl;
            return 0;
        }
    }

    for(int i=A;i<C;i++){
        if(i+1==C||i+2==C) break;
        if(S[i+1]=='.'&&(f||i+1!=D));
        else if(S[i+2]=='.'&&(f||i+2!=D)){
            i++;
        }
        else {
            cout<<"No"<<endl;
            return 0;
        }
    }



    cout<<"Yes"<<endl;


}