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>
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)
using namespace std;
using P = pair<int,int>;
typedef long long ll;
const ll mod = 1e9+7;
#define all(x) (x).begin(),(x).end()

template<class T> inline void chmin(T& a, T b) {
    if (a > b) {
        a = b;
    }
}

template<class T> inline void chmax(T& a, T b) {
    if (a < b) {
        a = b;
    }
}
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;
}

long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    std::cout<<std::fixed<<std::setprecision(10);
    ll n,a,b;
    cin >> n >> a >> b;
    ll nca = 1,ncb = 1;
    REP(i,a){
        nca *= (n-i)*modinv(i+1,mod)%mod;
        nca%=mod;
    }
    REP(i,b){
        ncb *= (n-i)*modinv(i+1,mod)%mod;
        ncb %= mod;
    }
    ll answer = modpow(2,n,mod) - ncb - nca - 1;
    answer %= mod;
    if(answer < 0)answer += mod;
    cout <<answer<< "\n";
}