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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif
#define int long long
const int MOD = 1e9 + 7;

struct Combination{
  const int MOD = 1e9+7;

  vector<int> fact; //　fact[i] = iの階乗
  void init(int n){
    fact.resize(n+1);
    fact[0] = fact[1] = 1;
    for(int i = 2; i <= n; i++){
      fact[i] = i * fact[i-1] % MOD;
    }
  }

  int nCr(int n, int r){ // nCr = n!/r!(n-r)!
    if(n < r) return 0;
    return fact[n] * mod_pow(fact[r]*fact[n-r]%MOD, MOD-2) % MOD;
  }

  int mod_pow(int n, int p){ // a/n ≡ a*n^(p-2) nとpは互いに素
    int r = 1;
    for(; p > 0; p >>= 1){
      if(p & 1LL) r = (r * n) % MOD;
      n = (n * n) % MOD;
    }
    return r; // r = n^p % MOD
  }

}comb;

signed main(){

  int n, a, b; cin >> n >> a >> b;

  comb.init(1e6);
  int ans = comb.mod_pow(2, n);

  int res = 1, x = n;
  for(int i = 1; i <= a; i++){
    res *= (x - i + 1) % MOD * comb.mod_pow(i, MOD-2) % MOD;
    res %= MOD;
  }
  //cout << res << endl;
  ans = (ans - res + MOD) % MOD;

  res = 1, x = n;
  for(int i = 1; i <= b; i++){
    res *= (x - i + 1) % MOD * comb.mod_pow(i, MOD-2) % MOD;
    res %= MOD;
    //cout << x-i+1 << " " << i << " " << res << endl;
  }
  //cout <<res << endl;
  ans = (ans - res + MOD) % MOD;

  if(n == 2) ans = 0;
  cout << (ans - 1) % MOD << endl;

  return 0;
}