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()
a -= 1
b -= 1
c -= 1
d -= 1

"""
Sa = "."
for i in range(a+1,len(S)) :
    Sa += S[i]
    if S[-2:] == "##" :
        break
for i in range(a-1,-1,-1) :
    Sa += S[i]
    if S[:2] == "##" :
        break

Sb = ""
for i in range(b+1,len(S)) :
    Sb += S[i]
    if S[-2:] == "##" :
        break
for i in range(b-1,-1,-1) :
    Sb += S[i]
    if S[:2] == "##" :
        break
"""
if S[a:(c+1)].count("##") == 0 :
    flg1 = 1
else :
    flg1 = 0

if S[b:(d+1)].count("##") == 0 :
    flg2 = 1
else :
    flg2 = 0

if c < d :
    flg3 = 1
else :
    if S[a:(c+1)].count("...") >= 1 :
        flg3 = 1
    else :
        flg3 = 0

    
if flg1*flg2*flg3 :
    print("Yes")
else :
    print("No")