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())
S = input()
dot = []
sharp = []
A_clear = True
B_clear = True

if C<D:
    # print(S[B-1:D])
    tmp_s = 0
    for s in S[B-1:D]:
        if s == '#' and tmp_s>0 :
            B_clear = False
            break
        elif s == '#':
            tmp_s += 1
        else:
            tmp_s = 0

    tmp_s = 0
    for s in S[A-1:C]:
        if s == '#' and tmp_s>0 :
            A_clear = False
            break
        elif s == '#':
            tmp_s += 1
        else: 
            tmp_s = 0
    if A_clear == True and B_clear == True:
        print('Yes')
    else:
        print('No')
    
elif D<C:
    # A
    tmp_s = 0
    for s in S[A-1:C]:
        if s == '#' and tmp_s>0 :
            A_clear = False
            break
        elif s == '#':
            tmp_s += 1
        else:
            tmp_s = 0
    # B
    tmp_s = 0
    tmp_d = 0
    A_clear = False
    for s in S[B-1:D]:
        if s == '.' and tmp_d>=2:
            tmp_s += 1
            A_clear = True
        elif s == '.' and tmp_s>=0:
            tmp_d += 1
            tmp_s = 0
        elif s == '#' and tmp_d>=0:
            tmp_s += 1
            tmp_d = 0
        elif s == '#' and tmp_s>0 :
            B_clear = False
            break
    if A_clear == True and B_clear == True:
        print('Yes')
    else:  
        print('No')