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;
int main()
{
    int n,a,b,c,d;
    string s;
    scanf("%d %d %d %d %d",&n,&a,&b,&c,&d);
    cin>>s;
    a--;
    b--;
    c--;
    d--;
    int x=0,y=0;
    int m=max(c,d);
    for(int i=min(a,b);i<m;i++)
    {
        if(s[i]=='#' && s[i+1]=='#')
        {
            x++;
            break;
        }
    }
    m=min(c,d);
    for(int i=min(a,b);i<=m-2;i++)
    {
        if(s[i]=='.' && s[i+1]=='.' && s[i+2]=='.')
        {
            y++;
            break;
        }
    }
    if(x)
    {
        printf("No");
        return 0;
    }
    if(d<c)
    {
        if(s[d-1]=='.' && s[d+1]=='.' && s[d]=='.')
        printf("Yes");
        else
        {
            if(y) printf("Yes");
            else printf("No"); 
        }
    }
    else printf("Yes") ;
    return 0;
}