Problem:
Score : 400 points

Problem Statement

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.)

Sample Input 1

4 1 3

Sample Output 1

7

In this case, Akari can choose 2 or 4 flowers to make the bouquet.

There are 6 ways to choose 2 out of the 4 flowers, and 1 way to choose 4, so there are a total of 7 different bouquets that Akari can make.

Sample Input 2

1000000000 141421 173205

Sample Output 2

34076506

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

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.*;
import java.math.*;

public class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int a = sc.nextInt();
    int b = sc.nextInt();
    sc.close();

    long MOD = 1000000007L;

    long sum = 0;
    long c = 1;
    long a_c = 0;
    long b_c = 0;
    for (int i = 1; i < n + 1; i++) {
      c *= modInv(i, MOD);
      c %= MOD;
      c *= (n - i + 1);
      c %= MOD;
      if (i == a) a_c = c;
      if (i == b) b_c = c;
      sum += c;
      sum %= MOD;
    }

    long ans = sum;
    ans -= a_c;
    if (ans < 0) ans += MOD;
    ans -= b_c;
    if (ans < 0) ans += MOD;
    System.out.println(ans);
  }

  private static long modInv(long x, long mod) {
    long a = x;
    long b = mod;
    long t = 1;
    long u = 0;
    while (0 < b) {
      long q = a / b;
      a -= q * b;
      { long tmp = b; b = a; a = tmp; }
      t -= q * u;
      { long tmp = u; u = t; t = tmp; }
    }

    if (t < 0) t += mod;
    return t;
  }
}