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: Accepted)

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

switch = 0

S_before_a = S.copy()
S_before_a[B-1] ='#'
A_list = S_before_a[A:C-1]
count1 = 0
for i in range (len(A_list)-1):
    if ''.join(A_list[i:i+2])=='##':
        count1+=1

S_after_a = S.copy()
S_after_a[C-1]='#'

count2 = 0
B_list = S_after_a[B:D-1]
for i in range (len(B_list)-1):
    if ''.join(B_list[i:i+2])=='##':
        count2+=1
    
#count逆
S_before_b = S.copy()
S_before_b[A-1] ='#'
B_list = S_before_b[B:D-1]
count3 = 0
for i in range (len(B_list)-1):
    if ''.join(B_list[i:i+2])=='##':
        count3+=1

S_after_b = S.copy()
S_after_b[D-1]='#'

count4 = 0
A_list = S_after_b[A:C-1]
for i in range (len(A_list)-1):
    if ''.join(A_list[i:i+2])=='##':
        count4+=1
        
change_counter =0
changer = S[max(A,B)-1:min(C,D)]
for i in range(len(changer)-2):
    if ''.join(changer[i:i+3])=='...':
        change_counter+=1
        
if (count1+count2) == 0 or (count3+count4) ==0:
    print('Yes')
elif (count1 == 1 and change_counter>0 ) or (count2 == 1 and change_counter>0 ) or (count3 == 1 and change_counter>0 ) or (count4 == 1 and change_counter>0 ):
    print('Yes')
else:
    print('No')