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: Runtime Error)

#include <stdio.h>

long long int factorial(long long int x)
{
  long long int temp = 1;
  while (x != 1 && x!= 0)
    temp *= x--;

  return temp;
}

int main(void)
{

  long long int n, cnt=0;
  long long int a, b, i;

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

  for (i = 1; i <= n; i++) {
    if (i != a && i != b) 
      cnt += factorial(n) / (factorial(i) * factorial(n-i));
    if (cnt >= 1000000000 + 7)
      cnt %= (1000000000 + 7);
  }

  printf("%lld", cnt);

  return 0;
}