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: Time Limit Exceeded)

#include <stdio.h>
/*--- 異なるn個からr個の整数を取り出す組み合わせの数を返す ---*/
int combination(int n, int r)
{
    if (r == 0 || r == n)
        return (1);
    else if (r == 1)
        return (n);
    return (combination(n - 1, r - 1) + combination(n - 1, r));
}
int main(){
	int N, a, b, sum;
	scanf("%d%d%d",&N,&a,&b);
	if (b - a != 1){
		for (int i = 1;i < a;i++){
			sum += combination(N, i) % 1000000007;
		}
		for (int i = a + 1;i < b;i++){
			sum += combination(N, i) % 1000000007;
		}
		for (int i = b + 1;i <= N;i++){
			sum += combination(N, i) % 1000000007;
		}
	}
	else{
		for (int i = 1; i < a;i++){
			sum += combination(N, i) % 1000000007;
		}
		for (int i = b + 1;i <= N;i++){
			sum += combination(N, i) % 1000000007;
		}
	}
	printf("%d",sum % 1000000007);
}