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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using ll = long long;
using itn = int;
using namespace std;
int GCD(int a, int b){
    return b ? GCD(b, a%b) : a;
}
int main(){
    int n,a,b,c,d;
    string s;
    cin >> n >> a >> b >> c >> d >> s;
    //まず、ふぬけ君がDまで到達できるかどうか
    for(int i=0; i<n-1; i++){
        if(s[i]=='#' && s[i+1]=='#'){
            puts("No");
            return 0;
        }
    }
    //次に、すぬけ君を動かす。
    //もし、素抜け君の最終地点がC<Dならさえぎるものがないので、ふぬけ君→すぬけ君の順にたどり着ける→YESの出力
    //D<Cだと厄介。ふぬけ君を先に通すと死ぬパターンがある。
    //Dの周囲±1に＃があるとすぬけ君がCまでたどり着かない。→NOの出力をいったん保留して以下を試す
    //この場合、「.」マスがBからDの間に３つあれば、ふぬけ君とすぬけ君の順番を変えられることを利用してみる（チェンジと呼ぶ）
    //チェンジしたはすぬけ君→ふぬけ君の順になるので奥のCに入り、Dで終えることができる→YESの出力
    //この判定がダメならNOを出力する
    bool change=false;
    if(c<d){
        puts("Yes");
    }else{
        for(int i=b-1; i<c-1; i++){
            if(s[i]=='.' && s[i+1]=='.' && s[i+2]=='.'){
                change=true;
            }
        }
        if(change) puts("Yes");
        else puts("No");
    }
    return 0;
}