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>
#include<math.h>
#include<stdlib.h>
#include<string.h>
#define max(x,y)(x>y?x:y)
#define min(x,y)(x>y?y:x)
#define MOD 1000000007
typedef long long int ll;
typedef unsigned long long int ull;
ll gcd(ll x,ll y){if(!y)return x;return gcd(y,x%y);}
ll lcm(ll x,ll y){return x/gcd(x,y)*y;}
void swap(ll *x,ll *y){ll tmp=*x;*x=*y;*y=tmp;}
int asc_cmp(const void* X,const void* Y){ll x=*(int*)X,y=*(int*)Y;if(x<y)return -1;else if(x==y)return 0;else return 1;}
int desc_cmp(const void* X,const void* Y){ll x=*(int*)X,y=*(int*)Y;if(x<y)return 1;else if(x==y)return 0;else return -1;}

ll modpow(ll a,ll n){
    ll res=1;
    while(n>0){
        if(n&1){res*=a;res%=MOD;}
        a*=a;a%=MOD;
        n>>=1;
    }
    return res;
}   
ll nCr(ll n,ll r){
    ll res,x,y;
    res=x=y=1;
    for(int i=0;i<r;i++){
        x*=(n-i);x%=MOD;
        y*=(r-i);y%=MOD;
    }
    res=modpow(y,MOD-2)%MOD;
    res=(res*x)%MOD;
    return res;
}


int main(){
    ll n,a,b;scanf("%lld %lld %lld",&n,&a,&b);

    ll x=modpow(2,n)-1;
    ll y=nCr(n,a)+nCr(n,b);
    y%=MOD;
    printf("%lld\n",modpow(2,n)-1-y);
    
      
}