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 Kotlin (1.0.0) (judge verdict: Wrong Answer)

fun main(args: Array<String>) {

    var (n ,a, b, c, d) = readLine()!!.split(" ").map { it.toInt() }
    a -= 1
    b -= 1
    c -= 1
    d -= 1
    val inputs = readLine()!!.toString()

    // 岩が連続してる場合 かつ一番遠いゴールがそれよりも先の場合
    if (inputs.indexOf("##") != -1 && inputs.indexOf("##") <= Math.max(c, d)) {
        println("No")
        return
    }

    // 岩が存在するが，すべてゴールより先にある場合 or 岩が存在しない場合
    if (inputs.indexOf("#") == -1 || inputs.indexOf("#") >= Math.max(c, d)) {
        println("Yes")
        return
    }

    // C ＜ D の場合 で B が先に D へ行ってしまえば良いので可能
    if (c < d) {
        println("Yes")
        return
    }

    /**
     * C > D でかつ，Dが岩の隣りにあって かつ 抜かせない場合
     */

    // C>D でかつ岩の隣に D がある(左右） かつ， A が Bを抜かせる場所がゴールよりも前にないとき
    // AをBより先に行かせられないとき

    fun isDNextToStone(): Boolean{
        if (d == inputs.lastIndex) {
            return inputs[d-1] == '#'
        }

        if (d == 0) {
            return inputs[d+1] == '#'
        }

        return inputs[d-1] == '#' || inputs[d+1] == '#'
    }

    fun canAovertakeB(): Boolean {
        if (inputs.subSequence(b-1, d).indexOf("...") == -1){
            return false
        }

        return true
    }

    // D の隣に岩がない場合
    if (inputs[d-1]=='.' && inputs[d+1]=='.') {
        println("Yes")
        return
    }

    // D の両側が岩の場合
    if (inputs[d-1]=='#' && inputs[d+1]=='#') {
        println("No")
        return
    }

    // C>D でかつ岩の隣に D がある(左右） かつ，A が B を抜かせる場所が B, D 内に無い
    if (isDNextToStone() &&  !canAovertakeB()) {
        println("No")
        return
    }

    //  D が岩の隣ない
    println("Yes")
}