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;
#define all(x) (x).begin(), (x).end()
typedef long long ll;

int main()
{
    int n, a, b, c, d;
    cin >> n >> a >> b >> c >> d;
    string s;
    cin >> s;
    int dame = 0;
    int kese=0;
    for(int i=a-1+1;i<c;i++){
        if(s[i]==s[i-1]&&s[i]=='#'){
            dame++;
        }
    }
    for(int i=b-1+1;i<d;i++){
        if(s[i]==s[i-1]&&s[i]=='#'){
            dame++;
        }
    }

    if(c>d){
        for(int i=b-1+2;i<d;i++){
            if('.'==s[i-1]&&'.'==s[i-2]&&s[i]=='.'){
                kese=-100000000;
            }
        }
        for(int i=b-1;i<d;i++){
            if(s[i]=='#'){
                kese++;
            }
        }
    }
    //cout<<dame<<" "<<kese<<endl;
    if(dame>0||kese>0){
        cout<<"No"<<endl;
    }
    else{
        cout<<"Yes"<<endl;
    }
}