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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <iostream>
#include <algorithm>
//#include <vector>

using namespace std;

#define MAX(x,y) (x>y?x:y)
#define MIN(x,y) (x<y?x:y)
#define ABS(x) (x>0?x:-(x))
#define INF 1000000000

#define MOD 1000000007

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

void swap2(double* a, double* b)
{
	double tmp;
	tmp = *a;
	*a = *b;
	*b = tmp;
}

int gcd(int a, int b)
{
	if (a < b)
		swap(&a, &b);

	if (a % b == 0)
		return b;
	else
		return gcd(b, a % b);
}

int lcm(int a, int b)
{
	return a * b / gcd(a, b);
}

int power(int a, int b)    //MODを法として、aのb乗を求める
{
	int p;

	if (b == 0) //0乗
		return 1;

	if (b % 2 == 0)
		return power((a * a) % MOD, b / 2);
	else
		return ((a * power(a, b - 1)) % MOD);
}

int choose(int n, int k) {
	return k == 0 ? 1 : (long long)choose(n - 1, k - 1) * n % MOD * power(k, MOD - 2) % MOD;
}

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

	scanf("%d%d%d", &n, &a, &b);
	ans = power(2, n) - 1 - choose(n, a) - choose(n, b);
	while (ans < 0)
		ans += MOD;
	printf("%d\n", ans);
	return 0;
}