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 5.4.1) (judge verdict: Runtime Error)

typedef long long int ll;

ll PowMod_RepeatSquaring(ll N, ll P, ll M){
	if (P == 0){
		return 1;
	}else if (P % 2 == 0){
		long int t = PowMod_RepeatSquaring(N, P/2, M); 
		return t * t % M;
	}else{
		return N * PowMod_RepeatSquaring(N, P - 1, M) % M;
	}
}  

ll comb(ll n, ll k, ll M){
	if (n < k) return 0;
	else if (n < 0 || k < 0) return 0;
  	else if (k == 0 || k == 1) return n;
  	if (k > n / 2) 	k = n - k;
    return comb(n, k - 1, M) * (n - k + 1) / k % M;
}

#include <stdio.h>
int main(){
  ll n, a, b, ans, M;
  M = 1000000007;
  scanf("%lld%lld%lld", &n, &a, &b);
  ans = (PowMod_RepeatSquaring(2, n, M) - comb(n, a, M) - comb(n, b, M) - 1) % M;
  printf("%lld", ans);
}