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)

#![allow(unused_imports)]

use std::{i32, i64, u32, u64, usize, f64};
use std::cmp::{max, min, Ordering};
use std::collections::*;
use std::io::{stdout, Write, BufWriter};
use std::mem::{swap};
use maplit::*;
use proconio::marker::{Isize1, Usize1, Bytes, Chars};

#[proconio::fastout]
fn main() {
    proconio::input!{
        _n: usize,
        a: Usize1,
        b: Usize1,
        c: Usize1,
        d: Usize1,
        s: Chars,
    }
    if d < c && ! s[b - 1..=d + 1].windows(3).any(|w| w.iter().copied().all(|s| s == '.')) {
        println!("No");
        return;
    }
    if s[a..=c].windows(2).any(|w| w.iter().copied().all(|s| s == '#')) && s[b..=d].windows(2).any(|w| w.iter().copied().all(|s| s == '#')) { println!("No"); }
    else { println!("Yes"); }
}