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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
#define ll long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
static const int MAX = 100000;
static const int NIL = -1;
using namespace std;

ll calC(ll a, ll b){

    if(a == 1) return 0;
    ll ans = 1;
    ll tmpB = b;
    ll tmpA = a;
    for(int i = 1; i <= b; i ++){
        ans *= tmpA;
        tmpA --;
    }
    for(int i = 1; i <= b; i ++){
        ans /= tmpB;
        tmpB --;
    }

    return ans;
}
int main() {
    int n,a,b,c,d,p; cin >> n >> a >> b >> c >> d;
    vector<char> ch(n+1,'.');
    string ans = "No";
    rep(i,n) cin >> ch[i];

    for(int i = b-1; i < d; i ++){
        if(ch[i-1] == '.' && ch[i] == '.' && ch[i+1] == '.' && i != c-1){
            ans = "Yes";
            p = i+1;
        }
    }
    if(c > p){
        if(ch[c-2] == '#' || ch[c] == '#') ans = "No";
    }
    for(int i = 0; i < n-1; i ++){
        if(ch[i] == '#' && ch[i+1] == '#') ans = "No";
    }
    cout << ans;


    return 0;
}