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 <sstream>
#include <vector>
#include <cassert>
#include <string>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <utility>
#include <tuple>
#include <regex>
#include <map>
#include <cstdint>
#include <set>
#include <deque>
#include <iomanip>
#include <queue>

using namespace std;
using ull = unsigned long long;
using ll = long long;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }

#define EPS (1e-7)
#define ALL(x) begin(x), end(x)

bool canGoal(const string& s, ll start, ll goal) {
	string sub = s.substr(start, goal - start);
	auto it = sub.find("##");
	return it == string::npos;
}

bool canCross(const string& s, ll start, ll goal) {
	string sub = s.substr(start-1, goal - start + 3);
	auto it = sub.find("...");
	return it != string::npos;
}

int main() {
	ll N, A, B, C, D;
	cin >> N >> A >> B >> C >> D;
	string S;
	cin >> S;

	// index計算のため0-origin
	A--;
	B--;
	C--;
	D--;

	bool ok = false;
	if (C < D) {
		// A C B D
		// A B C D
		// BがDにたどり着ける && AがCにたどり着ける
		ok = canGoal(S, B, D) && canGoal(S, A, C);
	} else {
		// A B D C
		// AがBを追い抜ける場所がある && AがCにたどり着ける &&BがDにたどり着ける(これは追い抜くので自明のような気もするが)
		ok = canCross(S, B, D) && canGoal(S, B, D) && canGoal(S, A, C);
	}
	const string ans = ok ? "Yes" : "No";
	cout << ans << endl;
	return 0;
}