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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
constexpr long long MOD = 1000000007;
long long mod_pow(long long x, long long n, long long mod) {// O(log(n))
    long long res = 1;
    while (n > 0) {//繰り返し二乗法
        if (n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}
//xの逆元を計算する
long long mod_inv(long long x, long long mod) {//O(log(mod))
    return mod_pow(x, mod - 2, mod);
}
long long find_numerator(long long n,long long a){
    long long p=1;
    for(long long i=0;i<=(a-1);i++){
        (p*=(n-i)) %=MOD;
    }
    return p;
}
long long find_denominator(long long n, long long a){
    long long q=1;
    for(long long i=1;i<=a;i++){
        (q*=i) %=MOD;
    }
    return q;
}
int main() {
long long n,a,b;
cin >> n >> a >> b;
long long all=mod_pow(2,n,MOD) - 1 + MOD;
all%=MOD;
long long c,d,e,f;
c=find_numerator(n,a);
d=find_denominator(n,a);
e=find_numerator(n,b);
f=find_denominator(n,b);
d=mod_inv(d,MOD);
f=mod_inv(f,MOD);
long long flowerofa=c*d%MOD;
long long flowerofb=e*f%MOD;
long long ans=(all - flowerofa - flowerofb + MOD + MOD)%MOD;
cout << ans << endl;
}