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 Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

Uses math;
var p:char;
    A,B,C,D,N,i,x,y:integer;
    S:array[0..200010] of integer;
begin
read(N,A,B,C,D);

for i := 1 to N do 
 begin 
  read(p); if ord(p) < 40 then S[i]:=1 else S[i]:=0; 
 end;

x:=1;
for i := 1 to max(C,D)-1 do 
 begin 
  if (S[i]=1) and (S[i+1]=1) then x:=0;
 end;
  
if x=0 then write('No')
 else
  begin
   if C > D then 
    begin
     x:=0;
     for i := B to N-2 do
      begin 
      if (S[i]=0) and (S[i+1]=0) and (S[i+2]=0) then x:=1;
      end;
    end;
   if C < D then x:=1;
    
    if x=0 then write('No') else write('Yes');
  end;
end.