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> 
#include <vector>
#include <algorithm>
#include <cmath>
#define fast ios_base::sync_with_stdio(0);cin.tie(0)
#define debug freopen("input", "r", stdin);freopen("output", "w", stdout)
#define endl '\n'
#define all(var) var.begin(), var.end()
typedef unsigned long long ll;
int const MOD = 1e9+7;
using namespace std;

ll fast_expo(ll a, ll n){
    ll res = 1;
    while(n){
        if(n&1)
            res = res*a%MOD;
        a=a*a%MOD;
        n >>= 1;
    }
    return res%MOD;
}

ll nCr(ll n, ll r){
    ll res1 = 1, res2 = 1;
    for(ll i = 1; i <= r; i++){
        res1 = res1*(n-i+1)%MOD;
        res2 = res2*i%MOD;
    }
    return res1*fast_expo(res2, MOD-2)%MOD;
}

int main(){
    ll n, a, b;
    cin >> n >> a >> b;
    ll ans = ((((fast_expo(2, n)-1-nCr(n,a)+MOD)%MOD) - nCr(n,b)+MOD)%MOD)%MOD;
    cout << ans << endl;
}