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 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#define DATALEN 200000

int main(void){
    int n;
    char s[DATALEN + 1];
    int a,b,c,d;
    int i,flag,max;
    int result = 1;//Yesなら1 Noなら0
    
    scanf("%d %d %d %d %d", &n, &a, &b, &c, &d);
    scanf("%s", s);
    max = (c > d) ? c : d;
    //2連の石判定
    for(i = 1; i <= max; i++){
        if(s[i] == '#' && s[i-1] == '#'){
            result = 0;
            break;
        }
    }
    
    //追い越し判定
    if(c > d){
        flag = 0;
        for(i = b + 1; i <= c; i++){
            if(s[i] == '.' && s[i-1] == '.' && s[i-2] == '.'){
                flag = 1;
                break;
            }
        }
        if(!flag){
            result = 0;
        }
    }
    
    if(!result){
        puts("No");
    }else{
        puts("Yes");
    }
    
    
    
    return 0;
}