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>
#define int long long
#define M 998244353
#define mod 1000000007
#define all(v) v.begin(),v.end()
#define ios ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define F first
#define S second
#define Zero(dp) memset(dp,0,sizeof(dp))
#define pb push_back
using namespace std;
int fast_power(int base,int power,int modulo)
{
	int result=1;
	while(power)
	{
		if(power%2)
		result=(result*base)%modulo;
		base=(base*base)%modulo;
		power/=2;
	}
	return result;
}
int inverse(int base,int modulo)
{
	return fast_power(base,modulo-2,modulo);
}
int32_t main(){
ios
int t=1;
//cin>>t;
while(t--)
{		

	int n,a,b;cin>>n;
	cin>>a>>b;
	int y1=1,y2=1,x1=1,x2=1;
	for(int i=a;i>0;i--)
		y1=(y1*i)%mod,x1=(x1*(n-(a-i)))%mod;
	for(int i=b;i>0;i--)
		y2=(y2*i)%mod,x2=(x2*(n-(b-i)))%mod;
	int ans1=fast_power(2,n,mod)-1;
	int ans2=(x1*(inverse(y1,mod)))%mod;
	int ans3=(x2*(inverse(y2,mod)))%mod;
	ans1=ans1-(ans2+ans3)+mod;
	ans1%=mod;
	cout<<ans1;
}
}