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>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <limits.h>

#define DIV 1000000007

long long cmb(long long n, long long r) {
	long long a = 1;
	long long min;

	if (n - r > r) min = r;
	else min = n - r;
	for (long long i = 0; i < min; i++)
		a = (a * ((n - i) / (i + 1))) % DIV;
	return a % DIV;
}

long long mypow(long long k, long long n) {
	long long ans = 1;
	int vec[1000] = { 0 };
	long long temp = 1;

	for (int i = 0; n > 0; i++) {
		vec[i] = n % 2;
		n /= 2;
	}
	temp = k;
	for (int i = 0; i < 500; i++) {
		//temp = temp * temp % DIV;
		if (vec[i]) ans = (ans * temp) % DIV;
		temp = temp * temp % DIV;
	}
	return ans % DIV;
}

int main(void) {
	long long n, a, b;
	long long ans ;

	scanf("%lld %lld %lld", &n, &a, &b);
	ans = (mypow(2, n)-1) % DIV;
	printf("%lld\n", ans);

	ans = (ans - cmb(n, a) % DIV) % DIV;
	if (ans < 0)ans = (ans + DIV) % DIV;
	printf("%lld\n", ans);

	ans = (ans - cmb(n, b) % DIV) % DIV;
	if (ans < 0)ans = (ans + DIV) % DIV;
	//printf("cmb(n,b)=%lld\n", cmb(n, b));

	printf("%lld", ans % DIV);
	return 0;
}