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;
using ll = long long;
using P = pair<int, int>;

const int MOD = 1000000007;

ll mod_pow(ll x, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

ll mod_inv(ll x, ll mod) { return mod_pow(x, mod - 2, mod); }

ll mod_comb(int n, int k, ll mod) {
    if (n < k) return 0;
    if (n - k < k) k = n - k;
    ll x = 1, y = 1;
    for (int i = 0; i < k; i++) x = x * (n - i) % mod;
    for (int i = k; i > 0; i--) y = y * i % mod;
    return x * mod_inv(y, mod) % mod;
}

int main() {
    int n, a, b;
    cin >> n >> a >> b;
    ll ans = (mod_pow(2, n, MOD) - 1) - mod_comb(n, a, MOD) - mod_comb(n, b, MOD);
    while (ans < 0) ans += MOD;
    cout << ans << endl;
}