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 <queue>
#include <set>
#include <map>
#include <utility>
#include <stack>
#include <numeric>
#include <algorithm>
#include <stdio.h>
#include <bitset>
#include <complex>
#include <iomanip>
using namespace std;
using ll = long long;
#define MM 1000000000
#define MOD MM+7
#define pi pair<int32,int32>
#define pl pair<int64,int64>
#define chmax(a,b) (a<b ? a=b:0)
#define chmin(a,b) (a>b ? a=b:0)
const long double PI = acos(-1);
const long long INF = 1e15;
int dx[8] = {-1,0,1,0,-1,-1,1,1};
int dy[8] = {0,-1,0,1,-1,1,1,-1};
template<typename T>
T GCD(T u,T v){
    return v ? GCD(v,u%v) : u;
}
int main(){
    int n,a,b,c,d; cin >> n >> a >> b >> c >> d;
    string s; cin >> s;
    s = "#" + s + "#";
    auto can_reach = [&](int start,int end){
        for(int i = start; i+1 <= end; i++){
            if (s[i] == '#' && s[i+1] == '#') return false;
        }
        return true;
    };
    if (!can_reach(a,c) || !can_reach(b,d)){
        cout << "No" << endl;
        return 0;
    }
    if (c > d){
        bool over = false;
        for(int i = b; i <= d; i++){
            if (s[i-1] == '.' && s[i] == '.' && s[i+1] == '.'){
                over = true;
            }
        }
        if (!over){
            cout << "No" << endl;
        }
    }
    cout << "Yes" << endl;
}