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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
    int n, a, b, i, j;
    scanf("%d%d%d", &n, &a, &b);
    long long num=1, A=1, B=1;
    for(i=0; i<n; i++) {
        num *= 2;
    }
    num -= 1;
    for(i=n; i>=n-a+1; i--) {
        for(j=1; j<=a; j++) {
            A /= j;
        }
        A *= i;
    }
    if(b > n/2) b = n - b;
    for(i=n; i>=n-b+1; i--) {
        for(j=1; j<=b; j++) {
            B /= j;
        }
        B *= i;
    }
    printf("%lld\n", num-A-B);
    return 0;
}