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: Time Limit Exceeded)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#define MOD 1000000007
using namespace std;
long long mfact(long long n,long long p)
{
    long long ans = 1;
    for(int i = 2;i <= n;i++)
    {
        ans = ans * i % p;
    }
    return ans;
}
long long mpow(long long x,long long n,long long p)
{
    long long ans = 1;
    while(n != 0)
    {
        if(n % 2 == 1)
        {
            ans = ans * x % p;
        }
            x = x * x % p;
            n = n / 2;
    }
    return ans;
}
long long mcomb(long long n,long long m,long long p)
{
    if(n == 0 && m == 1)
    {
        return 1;
    }
    if(n < m || n < 0)
    {
        return 0;
    }
    long long ans = mfact(n,p) * mpow(mfact(m,p),p - 2,p) % p;
    ans = ans * mpow(mfact(n - m,p),p - 2,p) % p;
    return ans;
}
long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
int main()
{
    long long int n,a,b;
    cin >> n >> a >> b;
    long long int total = modpow(2,n,MOD);
    total--;
    cout << (total - mcomb(n,a,MOD) - mcomb(n,b,MOD) + MOD) % MOD << "\n";

    return 0;
}