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<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<bitset>
#include<map>
#include<stack>
#include<queue>
#include<iomanip>
using namespace std;

#define repr(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(i, n) repr(i, 0, n)
#define INF 2e9
#define MOD 1e9+7
#define LINF (long long)4e18



int main(){
    int N,A,B,C,D;
    cin >> N >> A >> B >> C >> D;
    A--;B--;C--;D--;
    string S;
    cin >> S;
    if(C < D){
        bool ok = true;
        repr(i,A,N-1){
            if(S[i+1] == '#' && S[i] == '#') ok = false;
        }
        if(ok) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    else{
        bool ok = false;
        repr(i,B-1,D-1){
            if(S[i] == '.' && S[i+1] == '.' && S[i+2] == '.') ok = true;
        }
        repr(i,A,N-1){
            if(S[i+1] == '#' && S[i] == '#') ok = false;
        }
        if(ok) cout << "Yes" << endl;
        else cout << "No" << endl;

    }
}