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 Lua (5.3.2) (judge verdict: Accepted)

local function str2tbl(s)
    -- limit of #s is approximately 999959
    local t = {string.byte(s, 1, #s)}
    local char = string.char
    for i=1, #t do
        if char(t[i]) == '#' then
            t[i] = 1
        else
            t[i] = 0
        end
    end
    return t
end
--
local N, A, B, C, D = io.read("n", "n", "n", "n", "n", "l")
local S = str2tbl(io.read("l"))

local function YesNo(x)
    print(x and 'Yes' or 'No')
end

local function ok1(src, dst)
    local prev = S[src]
    for i=src+1, dst do
        local cur = S[i]
        if prev == 1 and cur == 1 then
            return false
        end
        prev = cur
    end
    return true
end

local function ok2(src, dst)
    for i=src, dst do
        local prev = S[i-1]
        local cur = S[i]
        local nex = S[i+1]
        if prev == 0 and cur == 0 and nex == 0 then
            return true
        end
    end
    return false
end

local fnuke_ok = ok1(B, D)
local snuke_ok = ok1(A, C)
if C < D then
    YesNo(fnuke_ok and snuke_ok)
else
    YesNo(ok2(B, D))
end