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: Wrong Answer)

#include <stdio.h>
#define MOD 1000000007
typedef long long ll;
ll modpow(ll a, ll n) {
	if (n == 0)
		return 1;
	if (n == 1)
		return (a % MOD);
	if (n % 2 == 1)
		return (a * modpow(a, n - 1) % MOD);
	ll t = modpow(a, n / 2);
	return (t * t % MOD);
}
ll comb(ll n, ll r) {
	ll res = 1;
	for (int i = n; i > n - r; i--) {
		res *= i;
		res %= MOD;
	}
	ll y = 1;
	for (int i = 2; i <= r; i++) {
		y *= i;
		y %= MOD;
	}
	y = modpow(y, MOD - 2);
	res = res * y % MOD;
	return res;
}
int main() {
	int n, a, b;
	ll ans = 1;
	scanf("%d%d%d", &n, &a, &b);
	ans = modpow(2, n) - 1 - comb(n, a) - comb(n, b);
	if (ans < 0) {
		ans += MOD;
	}
	printf("%lld\n", ans);
	return 0;
}