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 namespace std;
#define dbg(x) cerr << #x"=" << x << endl;
typedef long long LL;
const int MAX_N = 2*1e5+100;
char s[MAX_N];

int main(int argc, char const *argv[]){
	//freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(0); cin.tie(0);
    int N,A,B,C,D;
    cin >> N >> A >> B >> C >> D;
    cin >> (s+1);
    bool ok = true;
    for(int i = A+1; i <= C-2 && ok; ++i){
        if(s[i] == '#' && s[i+1] == '#') {ok = false; break;}
    }
    for(int i = B+1; i <= D-2 && ok; ++i){
        if(s[i] == '#' && s[i+1] == '#') {ok = false; break;}
    }
    if(ok && D < C){ // A C之间可达
        bool can = false;
        //s[A] = '#';
        for(int i = B+1; i <= D-1; ++i){
            if(s[i-1] == '.' && s[i] == '.' && s[i+1] == '.') {dbg(i); can = true; break;}
        }
        ok &= can;
    }
    cout << (ok ? "Yes" : "No") << endl;
	return 0;
}