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<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const ll mod = 1e9 +7;
const ll MAXN = 1e6 + 5;

ll p=mod;

inline ll qpow(ll a,ll b){
	ll base=a%p;
	ll ans=1;
	while(b){
		if(b&1) ans=(ans*base)%p;
		base=base*base%p;
		b>>=1;
	}
	return ans;
}

inline ll C(ll n,ll m){
	if(n<m) return 0;//组合数n<m特判
	if(m>n-m) m=n-m;//组合数性质
	ll a=1,b=1;
	for(int i=0;i<m;i++){
		a=(a*(n-i)%p);//组合数分子 a 
		b=(b*(i+1))%p;//组合数分母 b
	} 
	return a*qpow(b,p-2)%p;//费马小定理 a/b=a*b^(p-2) 
}

inline ll Lucas(ll n,ll m){
	if(m==0) return 1;
	return Lucas(n/p,m/p)*C(n%p,m%p)%p;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	ll n,a,b;
	cin>>n>>a>>b;
	cout<<(mod+qpow(2,n)-1-C(n,a)-C(n,b))%mod;
	return 0;
}