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: Accepted)

// ABC 156-D
// 2020.2.23 bal4u

#include <stdio.h>

typedef long long ll;

#define M 1000000007

int pow_mod(int b, int p) {
   int r = 1;
   while (p) {
      if (p & 1) r = (ll)r * b % M;
      b = (ll)b * b % M;
	  p >>= 1;
   }
   return r;
}

#define MOD 1000000007
#define MAX 200005
int fact[MAX] = {1,1};

int extended_gcd(int a, int b, int *x, int *y) {
	int d;
	if (b == 0) { *x = 1; *y = 0; return a; }
	d = extended_gcd(b, a % b, y, x);
	*y -= a / b * (*x);
	return d;
}

int inverse(int a) {
    int x, y;
    extended_gcd(a, M, &x, &y);
	return x % M;
}

int comb(int n, int a, int b) {
	int i, t, ans = 1;
	for (i = 0; i < a; ++i) ans = (ll)ans * (n-i) % M;
	t = 1;
	for (; i < b; ++i) t = (ll)t * (n-i) % M;
	return (ll)ans * (inverse(fact[a]) + (ll)t*inverse(fact[b])%M) % M;
}

int main()
{
	int i, n, a, b;
	ll ans;

	scanf("%d%d%d", &n, &a, &b);
	for (i = 2; i <= b; i++) fact[i] = (ll)fact[i-1]*i % M;
	ans = ((ll)pow_mod(2, n) - 1 - comb(n, a, b)) % M;
	if (ans < 0) ans += M;
	printf("%d\n", (int)ans);
	return 0;
}