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 Rust (1.42.0) (judge verdict: Wrong Answer)

// -*- coding:utf-8-unix -*-

use proconio::input;

fn main() {
    input! {
        n: usize,
        _a: u32,
        b: u32,
        c: u32,
        d: u32,
        s: String,
    }

    let vc: Vec<char> = s.chars().collect();
    if c == d {
        println!("No");
        return;
    } else if c > d {
        for i in b..d - 1 {
            if vc[i as usize] == '.' && vc[i as usize + 1] == '.' {
                println!("Yes");
                return;
            }
        }
        println!("No");
        return;
    }

    for i in 1..n {
        if vc[i] == '#' && vc[i - 1] == '#' {
            println!("No");
            return;
        }
    }

    println!("Yes");
}