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;
#define ll long long
const ll mod = 1e9 + 7;
ll n, a, b, res;
inline ll Pow (ll a, ll b) {
    ll t = 1;
    while (b) {
        if (b & 1) t = t * a % mod;
        a = a * a % mod, b >>= 1;
    } return t;
}
ll C (ll n, ll m) {
    if (n < m) return 0;
    ll res = 1;
    for (int i = 1; i <= m; i++) {
        ll a = (n + i - m) % mod;
        ll b = i % mod;
        res = res * (a * Pow (b, mod - 2) % mod) % mod;
    } return res;
}
ll Lucas (ll n, ll m) {
    if (m == 0) return 1;
    return C (n % mod, m % mod) * Lucas (n / mod, m / mod) % mod;
}
signed main() {
    scanf ("%lld %lld %lld", &n, &a, &b);
    res = Pow (2, n) - 1; ll tmp = Lucas (n, a) + Lucas (n, b);
    return printf ("%lld\n", ((res - tmp) % mod + mod) % mod), 0;
}