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 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<string.h>

int main()
{
    long long int n, answer = 1, d = 1;
    long long int a, b;
    long long int n_step = 1, a_step, n_a_step, b_step, n_b_step, n_c_a, n_c_b;

    for (int i = 0; i<9; i++)
    {
        d *= 10;
    }

    d += 7;

    scanf("%lld %lld %lld", &n, &a, &b);

    for (long long int i = 0; i<n; i++)
    {
        answer *= 2;

        if (answer > d)
        {
            answer %= d;
        }

        n_step *= (i + 1);

        if (n_step >d)
        {
            n_step %= d;
        }
    
        if ((i + 1) == a)
        {
            a_step = n_step;
        }
        
        if ((i + 1) == (n - a)){
            n_a_step = n_step;
        }
        
        if ((i + 1) == b){
            b_step = n_step;
        }
        
        if ((i + 1) == (n - b)){
            n_b_step = n_step;
        }
    
    }

    n_c_a = n_step;
    n_c_b = n_step;

    for (long long int i = 0; i<(d-2); i++)
    {
        n_c_a *=((a_step)*(n_a_step));
        n_c_b *=((b_step)*(n_b_step));

        if (n_c_a > d)
        {
            n_c_a %= d;
        }

        if (n_c_b > d)
        {
            n_c_b %= d;
        }
    }

    answer += 2*d;

    answer = answer - n_c_a - n_c_b; 

    answer += -1;

    answer %= d;

    printf("%lld", answer);

    return 0;
}