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 Python (3.4.3) (judge verdict: Wrong Answer)

N ,A, B, C, D = map(int, input().split())

rock = "#"
S = list()
three_empty = []
two_rocks = []
rock_cnt, empty_cnt = 0, 0

# 通れない２連続岩の場所とABが交差できる３emptyの場所取得
for i, s in enumerate(input().strip(), start=1):
  is_rock = s == rock
  if is_rock:
    if empty_cnt >= 3:
      three_empty.append(i - 1)
    rock_cnt += 1
    empty_cnt = 0
  else:
    if rock_cnt >= 2:
      two_rocks.append(i - 1)
    rock_cnt = 0
    empty_cnt += 1
if rock_cnt >= 2:
  two_rocks.append(i)
if empty_cnt >= 3:
  three_empty.append(i)

flag = True
left = A if A < B else B
right = B if A > B else A
left_rock = left
right_rock = 200001


for i in two_rocks:
  # 目標の間に岩があるか確認
  if A > i > C or B > i > D or A < i < C or B < i < D:
    flag = False
    break
  # 左と右にいる人から一番近い岩の位置
  if right_rock == 200001 and right < i:
    right_rock = i

if not flag:
  # 目標との間に2岩があるとき
  print("No")
elif (A > B and C > D) or (A < B and C < D):
  # 木彫との間に2岩がなく、交差が発生しない場合。
  print("Yes")
elif not three_empty:
  # 交差点となる3emptyがない場合
  print("No")
else:
  # 双方の一番近い岩との間に3emptyがあるかどうか 
  flag = False
  for i in three_empty:
    if left_rock < i < right_rock:
      flag = True
      break
  if flag:
    print("Yes")
  else:
    print("No")