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 <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
#define chmin(a, b) ((a) = min((a), (b)))
#define chmax(a, b) ((a) = max((a), (b)))
#define fs first
#define sc second
#define eb emplace_back
using namespace std;

typedef long long ll;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;

const ll MOD = 1e9 + 7;
const ll INF = 1e18;
const double pi = acos(-1);
const double eps = 1e-10;

int dx[] = {1, 0, -1, 0};
int dy[] = {0, -1, 0, 1};

int main(){
    int n, a, b, c, d; cin>>n>>a>>b>>c>>d;
    string s; cin>>s;
    bool valid = true;
    for(int i=a+1; i+1<c; i++){
        if(s[i] == '#' && s[i+1] == '#') valid = false;
    }

    for(int i=b+1; i+1<d; i++){
        if(s[i] == '#' && s[i+1] == '#') valid = false;
    }

    if(!valid){
        cout << "No" << endl;
    }
    else{
        if(a < b && b < c && c < d){
            cout << "Yes" << endl;
        }
        else if(a < c && c < b && b < d){
            cout << "Yes" << endl;
        }
        else{
            bool jump = false;
            for(int i=b-1; i+2<n && i+1<=d; i++){
                if(s[i] == '.' && s[i+1] == '.' && s[i+2] == '.'){
                    jump = true;
                }
            }

            if(jump){
                cout << "Yes" << endl;
            }
            else{
                cout << "No" << endl;
            }
        }
    }
}