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>
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
using namespace std;
typedef long long ll;
const ll p = 1e9+7;

ll modPow(int a, int n){
    if(n==1) return a % p;
    if(n%2 == 1) return (a * modPow(a, n-1)) % p;
    ll x = modPow(a, n/2);
    return (x * x) % p;
}

ll modCombine(ll a, ll b){
    b = min(b, a-b);
    ll y=1;
    for(ll i=1;i<=b;i++){
        y = (y*i) % p;
    }
    y = modPow(y,p-2);
    ll x = 1;
    for(ll i=a-b+1;i<=a;i++){
        x = (x*i) % p;
    }
    return (x * y) % p;
}

int main(){
    ll n,a,b;
    cin >> n >> a >> b;
    ll c1 = (modPow(2, n) - 1) % p;
    ll c2 = modCombine(n,a);
    ll c3 = modCombine(n,b);
    ll res = (c1 - c2 - c3 + p + p) % p;
    cout << res << endl;
    return 0;
}