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<vector>
#include<string>
#include<algorithm>
#include<tuple>
#include<map>
#include<queue>
#include<math.h>
#include<numeric>
#include<iomanip>
#define rep(i, n) rep2(i,0,n)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
#define PI acos(-1);
typedef long long ll;
using namespace std;
const int MOD=1000000007;
//scanf("%d", &a); '\n';


int main() {
	int n, a, b, c, d;
	cin >> n >> a >> b >> c >> d;
	vector<char> s(200002);
	rep(i, n) cin >> s[i];
	a -= 1;
	b -= 1;
	c -= 1;
	d -= 1;
	s[a] = 'A';
	s[b] = 'B';
	if (c < d) {
		while (true) {
			int tmpa = a;
			int tmpb = b;
			if (s[b + 2] == '.'&&b + 2 <= d) {
				b += 2;
				s[b] = 'B';
				s[b - 2] = '.';
			}
			if (s[b + 1] == '.'&&b + 1 <= d) {
				b += 1;
				s[b] = 'B';
				s[b - 1] = '.';
			}
			if (s[a + 2] == '.'&&a + 2 <= c) {
				a += 2;
				s[a] = 'A';
				s[a - 2] = '.';
			}
			if (s[a + 1] == '.'&&a + 1 <= c) {
				a += 1;
				s[a] = 'A';
				s[a - 1] = '.';
			}
			if (a == c && b == d) { cout << "Yes"; return 0; }
			if (a == tmpa && b == tmpb) { cout << "No"; return 0; }
		}
	}
	else{
		while (true) {
			int tmpa = a;
			int tmpb = b;
			if (s[a + 2] == '.'&&a + 2 <= c) {
				a += 2;
				s[a] = 'A';
				s[a - 2] = '.';
			}
			else if (s[a + 1] == '.'&&a + 1 <= c) {
				a += 1;
				s[a] = 'A';
				s[a - 1] = '.';
			}
			else if (s[b + 1] == '.'&&b + 1 <= d) {
				b += 1;
				s[b] = 'B';
				s[b - 1] = '.';
			}
			else if (s[b + 2] == '.'&&b + 2 <= d) {
				b += 2;
				s[b] = 'B';
				s[b - 2] = '.';
			}
			if (a == c && b == d) { cout << "Yes"; return 0; }
			if (a == tmpa && b == tmpb) { cout << "No"; return 0; }
		}
	}
	}