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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <iostream>

using namespace std;

// return x^n % mod
// It take O(log{n})
long long mod_pow(long long x, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    // 最下位bitが立っていたら x^{2^i}を掛ける
    if (n & 1) { res = res * x % mod; }
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}

// return x^{-1} % mod
// it take O(log{mod})
long long mod_inverse(long long x, long long mod) {
  return mod_pow(x, mod-2, mod);
}

// return nCk % mod
// It take O(k)
long long mod_comb(long long n, long long k, long long mod) {
  long long numer = 1, denom = 1;
  for (long long i = 0; i < k; i++) {
    // numer: 分子, denom: 分母
    numer = numer * ((n-i) % mod) % mod;
    denom = denom * ((i+1) % mod) % mod;
  }
  return numer * mod_inverse(denom, mod) % mod;
}

int main() {
  long long n, a, b;
  long long mod = 1000000007;

  cin >> n >> a >> b;

  long long ans = mod_pow(2, n, mod)-1;
  ans -= mod_comb(n, a, mod);
  ans %= mod;
  ans -= mod_comb(n, b, mod);
  ans %= mod;
  cout << (ans + mod) % mod << endl;

  return 0;
}