Problem: Sig has built his own keyboard. Designed for ultimate simplicity, this keyboard only has 3 keys on it: the `0` key, the `1` key and the backspace key.

To begin with, he is using a plain text editor with this keyboard. This editor always displays one string (possibly empty). Just after the editor is launched, this string is empty. When each key on the keyboard is pressed, the following changes occur to the string:

* The `0` key: a letter `0` will be inserted to the right of the string.
* The `1` key: a letter `1` will be inserted to the right of the string.
* The backspace key: if the string is empty, nothing happens. Otherwise, the rightmost letter of the string is deleted.



Sig has launched the editor, and pressed these keys N times in total. As a result, the editor displays a string s. Find the number of such ways to press the keys, modulo 10^9 + 7.

Constraints

* 1 ≦ N ≦ 5000
* 1 ≦ |s| ≦ N
* s consists of the letters `0` and `1`.

Input

The input is given from Standard Input in the following format:


N
s


Output

Print the number of the ways to press the keys N times in total such that the editor displays the string s in the end, modulo 10^9+7.

Examples

Input

3
0


Output

5


Input

300
1100100


Output

519054663


Input

5000
01000001011101000100001101101111011001000110010101110010000


Output

500886057

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	static final int P = 1_000_000_007;

	int[] goNoEmpty(int n, int k) {
		int[] ret = new int[n + 1];
		int[] dp = { 1 };
		int p2 = 1;
		for (int i = 1; i <= n; i++) {
			int[] nxt = new int[i + 1];
			for (int j = 1; j <= i; j++) {
				nxt[j] = (dp[j - 1] + (j + 1 < i ? dp[j + 1] : 0)) % P;
			}
			dp = nxt;
			if (i >= k && (i - k) % 2 == 0) {
				ret[i] = (int) ((long) dp[k] * p2 % P);
				p2 = 2 * p2 % P;
			}
		}
		return ret;
	}
	
	int[] goEmpty(int n) {
		int[] ret = new int[n + 1];
		ret[0] = 1;
		int[] dp = { 1 };
		for (int i = 1; i <= n; i++) {
			int[] nxt = new int[i + 1];
			nxt[0] = i == 1 ? 1 : (dp[0] + dp[1]) % P;
			for (int j = 1; j <= i; j++) {
				nxt[j] = (2 * dp[j - 1] % P + (j + 1 < i ? dp[j + 1] : 0)) % P;
			}
			dp = nxt;
			ret[i] = dp[0];
		}
		return ret;
	}

	void solve() throws IOException {
		int n = nextInt();
		int k = nextToken().length();

		int[] noEmpty = goNoEmpty(n, k);
		int[] empty = goEmpty(n);
		
		int ret = 0;
		for (int i = 0; i <= n; i++) {
			ret += (int)((long)noEmpty[i] * empty[n - i] % P);
			ret %= P;
		}
		
		out.println(ret);
	}

	Main() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new Main();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}