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: Runtime Error)

#include<bits/stdc++.h>
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
 
#define int long long
#define pii pair<int,int>
#define pqi priority_queue<int>
#define pb push_back()
#define INF 1000000000000000000
#define pi acos(-1)
#define deb(x) cout<<#x" = "<<x<<" "
#define ordered_set tree<int,null_type,less<int>,rb_tree_tag,tree_order_statistics_node_update>
const int modu=1e9+7;
 
int bigmod(int n,int p){
    if(p==0) return 1;
    int x=bigmod(n,p/2)%modu;
    x=(x*x)%modu;
    if(p%2) return (x*n)%modu;
    return x;
}
 
int modInverse(int a){
    return bigmod(a,modu-2)%modu;
}
 
void faster(){
    ios_base::sync_with_stdio(false);
    cin.tie();
    cout.tie();
}
 
void defInit(){
    faster();
    #ifdef OFFLINE
      auto dummy_in=freopen("in.txt", "r", stdin);
      auto dummy_out=freopen("out.txt", "w", stdout);
      if(0) dummy_in=dummy_out;
      else dummy_out=dummy_in;
    #endif
}

int add(int x,int y){
    int z =(x%modu + y%modu)%modu;
    while(z<0) z=(z+modu)%modu;
    return z;
}
 
int mul(int x,int y){
    int z = (x%modu * y%modu)%modu;
    while(z<0) z=(z+modu)%modu;
    return z;
}
// starts here

int32_t main(){
    faster();
    //Now I will start writing codes and start preparation to get errors
 	int n,a,b;
    cin>>n>>a>>b;
    int ans=bigmod(2LL,n)-1;
    int fact[n+1];
    fact[0]=1;
    for(int i=1;i<=n;i++){
        fact[i]=mul(fact[i-1],i);
    }
    int nca=mul(fact[n],mul(modInverse(fact[a]) , modInverse(fact[n-a]) ) );
    int ncb=mul(fact[n],mul(modInverse(fact[b]) , modInverse(fact[n-b]) ) );
    ans=add(ans,-add(nca,ncb));
    cout<<ans;
    return 0;

}