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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll =long long;
#define all(v) v.begin(),v.end()


ll com(ll n,ll a) {
vector<ll> vec(a);
  for(ll i=0;i<a;i++) {
vec[i]=n-i;
  }
  
  ll s=vec[0];
for(ll i=0;i<a;i++) {
ll k=s%(a-i);

    for(ll j=0;j<1000000000;j++) {
if(vec[k+(a-i)*j]%(a-i)==0) {
vec[k+(a-i)*j]/=(a-i);
break;
}
    }
  }
  
  ll ans=1;
  for(ll i=0;i<a;i++) {
ans*=vec[i];
    ans%=1000000007;
  }
  
  return ans;
}
 
int main() {
  ll n,a,b;
  cin>>n>>a>>b;
  ll INF=1000000007;
  ll L=1;
  for(ll i=0;i<1000;i++) {
    L*=2;
  L%=INF;
  }
  ll k=n/1000;
  n%=1000;
  ll A=1;
 for(ll i=0;i<n;i++) {
   A*=2;
 A%=INF;
 }
  for(ll i=0;i<k;i++) {
A*=L;
    A%=INF;
  }

  
  A--;
 A-=com(n,a);
  A-=com(n,b);
  if(A<0) {
while(A<0) {
A+=INF;
}
  }
  

  cout<<A<<endl;
}