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.

Examples

Input

7 1 3 6 7
.#..#..


Output

Yes


Input

7 1 3 7 6
.#..#..


Output

No


Input

15 1 3 15 13
...#.#...#.#...


Output

Yes
import java.io.PrintStream;
import java.util.Scanner;

public class Main {
	static final PrintStream out = System.out;
	static Scanner sc = new Scanner(System.in);
	static final long Mod = 1000_000_007L;
	public static void main(String[] args) {
		Main main = new Main();
		main.solve();
		System.exit(0);
	}

	void solve() {
		int N = nextInt();
		int A = nextInt();//here
		int B = nextInt();//here
		int AG = nextInt();//目的地
		int BG = nextInt();//目的地
		String S = next();
		//char[] cArray = S.toCharArray();
		String res = "No";

		if(N>1) {
			//移動範囲
			String tS = S.substring(Math.min(A, B), Math.max(AG,BG));
			//##が連続している場合達成不可能である
			if(tS.replace("##", "-").length()!=tS.length()) {
				out.println(res);
				return;
			}
		}

		boolean overtakef = false;
		//追い越す必要があるならtrue
		if(A < B && AG > BG || A > B && AG < BG) overtakef = true;
		if(overtakef) {
			String tS = S.substring(Math.max(A, B)-2, Math.min(AG,BG)+1);

			//out.println(tS.replace("...", "-")+" "+tS.length());
			//...が存在しない場合追い越すことができず達成不可能である
			if(tS.replace("...", "-").length()==tS.length()) {
				out.println(res);
				return;
			}
		}
		out.println("Yes");
	}
	private long max(long a, long b) {
		return Math.max(a, b);
	}
	private long min(long a, long b) {
		return Math.min(a, b);
	}
	private static int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)a[i] = nextInt();
		return a;
	}
	private static long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)a[i] = nextLong();
		return a;
	}
	private static int nextInt() {
		return Integer.parseInt(next());
	}
	private static long nextLong() {
		return Long.parseLong(next());
	}
	private static String next() {
		return sc.next();
	}
	static long gcd(long m,long n) {
		if (m<n) return gcd(n,m);
		if (n==0)return m;
		return gcd(n,m%n);
	}

}

/*
8 1 3 7 4
.#..#...
No

4 1 2 4 3
....
Yes

6 1 2 6 5
..#...
Yes
*/