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: Accepted)

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

long long combi(long long,long long);
long long modpow(long long,long long);

int main(){
    long long n,a,b,sum=1,i,mod=1000000007;scanf("%lld%lld%lld",&n,&a,&b);
    sum=modpow(2,n);//2^(10^9)=140625001
    sum--;//combinationの和なので2の累乗を用いる
    
    sum=sum-combi(n,a);//嫌いな数になる場合を除く
    sum=sum-combi(n,b);
    
    sum=sum%mod;

    if(sum>=0){
        printf("%lld\n",sum);
    }else{
        printf("%lld\n",sum+mod);
    }
    return 0;
}

long long combi(long long n,long long r){//繰り返し二乗法を用いてcombination計算
    long long ans,i,under=1,up=1,mod=1000000007;
    for(i=1;i<=r;i++){
        under=under*i;
        under=under%mod;
    }
    under=modpow(under,mod-2);
    
    for(i=1;i<=r;i++){
        up=up*(n-i+1);
        up=up%mod;
    }
    ans=up*(under%mod);
    return ans%mod;
}

long long modpow(long long a,long long n){//繰り返し二乗法
    long long mod=1000000007;
    if(n==1) return a%mod;
    if(n%2==1) return (a*modpow(a,n-1))%mod;
    long long t=modpow(a,n/2);
    return (t*t)%mod;
}