Akari has n kinds of flowers, one of each kind.

She is going to choose one or more of these flowers to make a bouquet.

However, she hates two numbers a and b, so the number of flowers in the bouquet cannot be a or b.

How many different bouquets are there that Akari can make?

Find the count modulo (10^9 + 7).

Here, two bouquets are considered different when there is a flower that is used in one of the bouquets but not in the other bouquet.

Constraints

* All values in input are integers.
* 2 \leq n \leq 10^9
* 1 \leq a < b \leq \textrm{min}(n, 2 \times 10^5)

Input

Input is given from Standard Input in the following format:


n a b


Output

Print the number of bouquets that Akari can make, modulo (10^9 + 7). (If there are no such bouquets, print `0`.)

Examples

Input

4 1 3


Output

7


Input

1000000000 141421 173205


Output

34076506
import java.util.*;
import java.io.*;

public class Main {
	static int mod = (int) 1e9 + 7;
	static int DX[] = { -1, 0, 1, 0 }, DY[] = { 0, -1, 0, 1 };

	public static void main(String[] args) {
		FastScanner fs = new FastScanner(System.in);
		int n = fs.nextInt(), a = fs.nextInt(), b = fs.nextInt();
		long ans = modpow(2,n);
		long s = fermatComb(n, a); 
		long t = fermatComb(n, b);
		ans = sub(ans,s);
		ans = sub(ans,t);
		ans = sub(ans,1);
		System.out.println(ans);
	}
	
	//MOD culculations
	static long plus(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x + y) % mod;
		return res;
	}

	static long sub(long x, long y) {
		x %= mod;
		y %= mod;
		long res = (x - y + mod) % mod;
		return res;
	}

	static long mul(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * y % mod;
		return res;
	}

	static long div(long x, long y) {
		x %= mod;
		y %= mod;
		long res = x * modpow(y, mod - 2) % mod;
		return res;
	}


	
	//Fermat Combination
	static long fermatComb(long a, long b) {
		if (b > a - b)
			return fermatComb(a, a - b);
		long ansMul = 1, ansDiv = 1;
		for (int i = 0; i < b; i++) {
			ansMul *= (a - i);
			ansMul %= mod;
			ansDiv *= (i + 1);
			ansDiv %= mod;
		}
		long ans = ansMul * modpow(ansDiv, mod - 2) % mod;
		return ans;
	}

	//modの元での高速pow()
	static long modpow(long a, long p) {
		if (p == 0)
			return 1;
		if (p % 2 == 0) {
			long halfP = p / 2;
			long root = modpow(a, halfP);
			return root * root % mod;
		} else
			return a * modpow(a, p - 1) % mod;
	}

}

//高速なScanner
class FastScanner {
	private BufferedReader reader = null;
	private StringTokenizer tokenizer = null;

	public FastScanner(InputStream in) {
		reader = new BufferedReader(new InputStreamReader(in));
		tokenizer = null;
	}

	public String next() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				tokenizer = new StringTokenizer(reader.readLine());
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken();
	}

	public String nextLine() {
		if (tokenizer == null || !tokenizer.hasMoreTokens()) {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
		}
		return tokenizer.nextToken("\n");
	}

	public long nextLong() {
		return Long.parseLong(next());
	}

	public int nextInt() {
		return Integer.parseInt(next());
	}

	public double nextDouble() {
		return Double.parseDouble(next());
	}

	public int[] nextIntArray(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = nextInt();
		return a;
	}

	public long[] nextLongArray(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nextLong();
		return a;
	}
}