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 <iostream>
using namespace std;
typedef long long LL;
const LL p = 1e9 + 7;

LL pow(LL a, LL b, LL p) {
    LL ans = 1;
    a %= p;
    while (b) {
        if (b & 1) {
            ans = (ans % p) * (a % p) % p;
        }
        b /= 2;
        a = (a % p) * (a % p) % p;
    }
    ans %= p;
    return ans;
}

LL inv(LL x, LL p) {
    return pow(x, p - 2, p);
}

LL C(LL n, LL m, LL p) {
    if (m > n) {
        return 0;
    }
    LL up = 1, down = 1;
    for (int i = n - m + 1; i <= n; i++) {
        up = up * i % p;
    }
    for (int i = 1; i <= m; i++) {
        down = down * i % p;
    }
    return up * inv(down, p) % p;
}

LL Lucas(LL n, LL m, LL p) {
    if (m == 0) {
        return 1;
    }
    return C(n % p, m % p, p) * Lucas(n / p, m / p, p) % p;
}

int main() {
    LL n, a, b, ans = 0;
    cin >> n >> a >> b;
    LL cnt = pow(2, n, p);
    ans = (cnt - 1 - C(n, a, p) - C(n, b, p) + (p << 1)) % p;
    //ans = (cnt - 1 - Lucas(n, a, p) - Lucas(n, b, p) + (p << 1)) % p;
    cout << ans << endl;
    return 0;
}