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)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int acs(const void *a, const void *b) { 
  return *(int*)a - *(int*)b;
} /* 1,2,3,4.. */
int des(const void *a, const void *b) {
  return *(int*)b - *(int*)a;
} /* 8,7,6,5.. */
int cmp_char(const void *a, const void *b) { 
  return *(char*)a - *(char*)b;
} /* a,b,c,d.. */
int cmp_str(const void *a, const void *b) {
  return strcmp(*(const char **)a, *(const char **)b);
} /* aaa,aab.. */
#define min(a,b) (a < b ? a : b)
#define max(a,b) (a > b ? a : b)
#define rep(i, l, r) for (int i = l; i < r; i++)
#define MAX 100001
#define MOD 1000000007
#define INF 1000000009

typedef long long int lli;

void swap(lli *a, lli *b)
{
  lli tmp;
  tmp = *a;
  *a = *b;
  *b = tmp;
}

// a ÷b (mod MOD) →(a % MOD) * modinv(b, MOD) % MDO
lli modinv(lli a, lli m) {
    lli b = m, u = 1, v = 0;
    while (b) {
        lli t = a / b;
        a -= t * b; swap(&a, &b);
        u -= t * v; swap(&u, &v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}


long long int combi(long long int n, long long int r) {
	long long int ans = 1;
	for ( long long int i = 0; i < r; i++ ) {
		ans = (ans * (n - i) % MOD) * modinv(r - i, MOD) % MOD;
	}
	return ans;
}

int main(void) {
	long long int n, a, b;
	scanf("%lld %lld %lld", &n, &a, &b);

	long long int sum = 2;
	long long int m = 1;
	while ( m < n ) {
		if ( m * 2 <= n ) {
			sum = sum * sum % MOD;
			m *= 2;
		} else {
			sum = sum * 2 % MOD;
			m++;
		}
	}

	sum = ((sum - combi(n, a) + MOD) % MOD - combi(n, b) - 1 + MOD) % MOD;
	printf("%lld\n", sum);
  return 0;
}