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 JavaScript (node.js v5.12) (judge verdict: Wrong Answer)

var lines = [];
var readline = require('readline');

var rl = readline.createInterface({
    input: process.stdin,
    output: process.stdout,
});

rl.on('line', function (x) {
    lines.push(x);
});

rl.on('close', function () {
    var N = Number(lines[0].split(" ")[0]);
    var A = Number(lines[0].split(" ")[1]);
    var B = Number(lines[0].split(" ")[2]);
    var C = Number(lines[0].split(" ")[3]);
    var D = Number(lines[0].split(" ")[4]);

    var S = lines[1].split("");

    // NOとなる場合，
    // ##がある
    // #Bがある
    var flag = false;
    for (var i = 0; i < S.length; i++) {
        if ((S[i] === "#" && S[i + 1] === "#") && i + 1 <= Math.max(C, D) && A <= i) { console.log("No"); return (0) }
        if ((S[i] === "." && S[i + 1] === "." && S[i + 2] === ".") && i + 2 <= Math.min(C, D) && B <= i) { flag = true }
    }
    if (D < C && (S[D - 2] === "#" || S[D] === "#") && (S[B - 2] === "#" || S[B] === "#") && flag === false) { console.log("No"); return (0) }

    console.log("Yes");
});