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: Wrong Answer)

#include<stdio.h>

#define modn 1000000007


static inline int mod_multi(long long a,long long b){
    unsigned long long ans;
    ans = (long long)a * b;
    if(ans > modn)ans %= modn;
    return ans; 
}

static inline int modpow(int number,int index){
    int ret = 1;
    while(index > 0){
        if(index & 1)ret = mod_multi(ret,number);
        number = mod_multi(number,number);
        index>>=1;
    }
    return ret;
}


static inline int inv(int n){
  return modpow(n,modn - 2);
}

int modcomb(long long n,int k){
  int ret = 1;
  if(k > n/2)k = n - k;
  for(long long i = n;n - i < k;i--)ret = mod_multi(i,ret);
  for(int i = 1;i <= k;i++)ret = mod_multi(ret,inv(i));
  return ret;
}

int main(void){
    int n,ans,a,b;
    scanf("%d %d %d",&n,&a,&b);
    ans = modpow(2,n);
    ans--;
    ans -= modcomb(n,a);
    ans -= modcomb(n,b);
    if(ans < 0)ans += modn;
    printf("%d",ans);
    return 0;
}