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<cstdio>
#include<iostream>
using namespace std;
#define ll long long
ll n,m,p;
ll pow_m(ll a,ll k,ll p)
{
    ll ans=1;
    ll tmp=a%p;
    while(k)
    {
        if(k&1)ans=ans*tmp%p;
        tmp=tmp*tmp%p;
        k>>=1;
    }
    return ans;
}
ll C(ll n,ll m,ll p)
{
    if(m>n)return 0;
    ll a=1,b=1;
    for(int i=1; i<=m; i++)
    {
        a=a*(n+i-m)%p;
        b=b*i%p;
    }
    return a*pow_m(b,p-2,p)%p;
}
ll Lucas(ll n,ll m,ll p)
{
    ll ans=1;
    while(n&&m)
    {
        ans=ans*C(n%p,m%p,p)%p;
        n/=p;
        m/=p;
    }
    return ans;
}
int main()
{
    ll n, a,b, p=1e9+7;
    cin>>n>>a>>b;
    cout<<(pow_m(2,n,p)-1-Lucas(n,a,p)-Lucas(n,b,p)+p)%p;
    return 0;
}