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.*;

public class Main {

    public static void main(String[] args) {
        // write your code here
        Scanner sc = new Scanner(System.in);

        int n = sc.nextInt();
        int as = sc.nextInt();
        int bs = sc.nextInt();
        int mod = 1000000007;

        long ans = pow(2, n, mod) - 1;
        long a = 1;
        for (int i = 2; i <= as; i++) {
            a *= i;
            a %= mod;
        }
        a = pow(a, mod - 2, mod);
        long ax = 1;
        for (int i = n; i >= n - as + 1; i--) {
            ax *= i;
            ax %= mod;
        }
        long minus = ax * a % mod;
        long b = 1;
        for (int i = 2; i <= bs; i++) {
            b *= i;
            b %= mod;
        }
        b = pow(b, mod - 2, mod);
        long bx = 1;
        for (int i = n; i >= n - bs + 1; i--) {
            bx *= i;
            bx %= mod;
        }
        minus += bx * b % mod;
        minus %= mod;

        System.out.println(ans - minus < 0 ? ans - minus + mod : ans - minus);
    }

    //return a^b mod M O(logB)
    private static long pow(long a, long b, int M) {
        long ret = 1;
        long tmp = a;
        while (b > 0) {
            if ((b & 1) == 1) ret = (ret * tmp) % M;
            tmp = (tmp * tmp) % M;
            b = b >> 1;
        }
        return ret;
    }
}
