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: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <map>
#include <queue> 
#include <stack>
#include <set>

using namespace std;
typedef long long ll;
#define rep(i, s, e) for (int(i) = (s); (i) <= (e); ++(i))
#define all(x) x.begin(),x.end()


int main()
{
	int n, a, b, c, d; cin >> n >> a >> b >> c >> d;
	--a, --b, --c, --d;
	string s; cin >> s;
	bool ok = true;
	int left = min(a, b), right = max(c, d);
	rep(i, left, right - 1)
	{
		if (s[i] == '#'&&s[i + 1] == '#')ok = false;
	}
	if (!ok)
	{
		cout << "No" << endl;
		return 0;
	}

	if (c < d) cout << "Yes" << endl;
	else
	{
		ok = false;
		rep(i, b, d)
		{
			if (s[i - 1] == '.'&&s[i] == '.'&&s[i + 1] == '.')ok = true;
		}
		if (ok)cout << "Yes" << endl;
		else cout << "No" << endl;
	}	
}