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;
const int MOD = 1e9 + 7;
typedef long long ll;
ll mod_pow(ll x, ll n, ll mod){//x^nのmodのmodを計算する
    if(n == 0) return 1;
    if(n % 2 == 0){
        ll t = mod_pow(x, n / 2, mod); //nが偶数のときはx^n = x^(n / 2) * x^(n / 2)
        return t * t % mod;
    } 
    return x * mod_pow(x, n - 1, mod) % mod;//nが奇数のときはx^n = x * x^(n - 1)
}

int main(){
    ll n;
    ll a, b;
    cin >> n >> a >> b;
    ll ans = mod_pow(2, n, MOD);
    ll finv[b];
    finv[0] = 1;
    ans--;
    for(int i = 0; i < b; i++){
        finv[i] = mod_pow(i + 1, MOD - 2, MOD);
    }
    ll na = 1, nb = 1;
    for(int i = 0; i < b; i++){
        nb = (nb * (n - i) % MOD) * finv[i] % MOD;
    }
    for(int i = 0; i < a; i++){
        na = (na * (n - i) % MOD) * finv[i] % MOD;
    }
    ans -= na + nb;
    if(ans < 0) cout << ans + MOD << endl;
    else cout << ans << endl;
}