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 PyPy3 (2.4.0) (judge verdict: Time Limit Exceeded)

###template###
import sys
def input(): return sys.stdin.readline().rstrip()
def mi(): return map(int, input().split())
###template###

N, A, B, C, D = mi()
#S[0~N-1]でアクセス
cells = input()

if C > D:
  c = C
  C = D
  D = c
  a = A
  A = B
  B = a

#全部0~N-1ベースにする
A -= 1
B -= 1
C -= 1
D -= 1

#既にスタック済みの座標の組(visited[0~N-1][0~N-1]でアクセス)
visited = [[0]*N for _ in range(N)]

from collections import deque
st = deque(maxlen=10000)
#初期座標
st.append((A,B))

#結果が出るまでループする
#スタックから座標の組を取り出し、F>Sの優先順位で動かしてスタックに入れる
num = 0
while num < 10000:
  num += 1
  #選択肢がなくなったら、駄目だったと分かる
  try:
    s, f = st.popleft()
  except:
    print('No')
    exit()

  #ここがゴールかどうか
  if s==C and f==D:
    print('Yes')
    exit()
  #Fが動けるかどうか
  fok = 0
  try:
    if visited[s][f+2]==0 and cells[f+2]=='.' and s != f+2:
      st.append((s,f+2))
      visited[s][f+2] = 1
      fok = 1
  except:
    pass
  try:
    if visited[s][f+1]==0 and cells[f+1]=='.' and s != f+1:
      st.append((s,f+1))
      visited[s][f+1] = 1
      fok = 1
  except:
    pass
  #Fが動けなかった場合のみSが動く
  if fok == 0:
    try:
      if visited[s+2][f]==0 and cells[s+2]=='.' and s+2 != f:
        st.append((s+2,f))
        visited[s+2][f] = 1
        fok = 1
    except:
      pass
    try:
      if visited[s+1][f]==0 and cells[s+1]=='.' and s+1 != f:
        st.append((s+1,f))
        visited[s+1][f] = 1
        fok = 1
    except:
      pass