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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll ZERO = 0;
const ll INF = 10000000000000000; //10^16
const ll MOD = 1000000007; //10^9+7

const ll MAX = 200003;


ll pow2(ll m){
    ll ret = 1;
    ll a = 2; //a=2^i
    while( m >= 1 ){
        if(m & 1){ //2進展開して下からi桁目が1ならaをかける
            ret = ret * a % MOD;
        }
        a = a * a % MOD;
        m >>= 1;
    }
    return ret;
}

int main() {
    vector<ll> inv(MAX);
    inv.at(1) = 1;
    for(int i = 2; i < 200003; i++){
        inv.at(i) = MOD - inv.at(MOD % i) * (MOD / i) % MOD;
    }
    
    ll n,a,b;
    cin >> n >> a >> b;
    
    ll ans = pow2(n);
    ll nca,ncb;
    nca = 1;
    ncb = 1;
    
    for(ll i = 1; i <= a; i++){
        nca = nca * (n-i+1) % MOD;
        nca = nca * inv.at(i) % MOD;
    }
    for(ll i = 1; i <= b; i++){
        ncb = ncb * (n-i+1) % MOD;
        ncb = ncb * inv.at(i) % MOD;      
    }
    
    ans -= nca;
    ans -= ncb;
    ans -= 1;
    ans = (ans + 3*MOD) % MOD;
    
    printf("%lld\n", ans);
}