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: Accepted)

import java.util.*;

class Main {
  
  public static final int mod = 1000000007;
  public static final int fmax = 200005;
  public static long fac[] = new long[fmax];
  public static long inv[] = new long[fmax];
  public static long finv[] = new long[fmax];
  
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    long n = Long.parseLong(sc.next());
    long a = Long.parseLong(sc.next());
    long b = Long.parseLong(sc.next());

    if(n==2 & a==1 & b==2){
      System.out.println("0");
      return;
    }
    
    NCRinit();
    
    long ap = n % mod;
    long bp = n % mod;
    
    for(long i=1; i<=a-1; i++){
      ap *= (n-i);
      ap %= mod;
    }
    for(long i=1; i<=b-1; i++){
      bp *= (n-i);
      bp %= mod;
    }
    
    for(int i=1; i<=a; i++){
      ap *= inv[i];
      ap %= mod;
    }
    for(int i=1; i<=b; i++){
      bp *= inv[i];
      bp %= mod;
    }
    
    long ans = modpow(2, (int)n);
    ans %= mod;
    
    ans -= 1L;
    ans %= mod;
    if(ans<0) ans += mod;
    ans -= ap;
    ans %= mod;
    if(ans<0) ans += mod;
    ans -= bp;
    ans %= mod;
    if(ans<0) ans += mod;
    
    System.out.println(ans);
    
  }

  // a^n mod を計算する
  public static long modpow(long x, int n) {
      long ret = 1;
      while (n > 0) {
          if ((n & 1) == 1) {
              ret = (ret * x)%mod;
          }
          x = (x * x)%mod;
          n >>= 1;
      }
      return ret;
  }

  // テーブルを作る前処理
  public static void NCRinit() {
    fac[0] = 1;
    fac[1] = 1;
    finv[0] = 1;
    finv[1] = 1;
    inv[0] = 1;
    inv[1] = 1;
    for (int i = 2; i < fmax; i++){
      fac[i] = fac[i - 1] * i % mod;
      inv[i] = mod - inv[mod%i] * (mod / i) % mod;
      finv[i] = finv[i - 1] * inv[i] % mod;
    }
  }
 
  // 二項係数計算
  public static long ncr(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
  }
  
}