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 Ruby (2.3.3) (judge verdict: Wrong Answer)

# frozen_string_literal: true

n, a, b, c, d = gets.split.map(&:to_i)
s = gets.chomp
a -= 1
b -= 1
c -= 1
d -= 1

oikoshi = false

if c < d
  oikoshi = true
else
  if s[b - 1] != '#' && s[b + 1] != '#'
    oikoshi = true
  else
    b.upto(c) do |i|
      if s[i] != '#' && s[i - 1] != '#' && s[i + 1] != '#'
        oikoshi = true
      end
    end
  end
end
dp1 = Array.new(n) { false }
dp2 = Array.new(n) { false }

dp1[a] = true
dp2[b] = true

dp1[0] = a.zero? ? true : false
dp1[1] = a.zero? || a == 1 ? true : false
dp2[0] = b.zero? ? true : false
dp2[1] = b.zero? || a == 1 ? true : false
0.upto(1) do |i|
  dp1[i] = dp2[i] = false if s[i] == '#'
end

[(a + 1), 2].max.upto(n - 1) do |i|
  dp1[i] = (s[i] == '#' ? false : (dp1[i - 1] || dp1[i - 2]))
end

[(b + 1), 2].max.upto(n - 1) do |i|
  dp2[i] = (s[i] == '#' ? false : (dp2[i - 1] || dp2[i - 2]))
end
if dp1[c] && dp2[d] && oikoshi
  puts 'Yes'
else
  puts 'No'
end