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;

const int MOD = 1e9 + 7;

class mint {
    long long x;
public:
    mint(long long x=0) : x((x % MOD + MOD) % MOD) {}
    mint operator-() const { 
      return mint(-x);
    }
    mint& operator+=(const mint& a) {
        if ((x += a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator-=(const mint& a) {
        if ((x += MOD-a.x) >= MOD) x -= MOD;
        return *this;
    }
    mint& operator*=(const  mint& a) {
        (x *= a.x) %= MOD;
        return *this;
    }
    mint operator+(const mint& a) const {
        mint res(*this);
        return res+=a;
    }
    mint operator-(const mint& a) const {
        mint res(*this);
        return res-=a;
    }
    mint operator*(const mint& a) const {
        mint res(*this);
        return res*=a;
    }
    mint pow(long long t) const {
        if (!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if (t&1) a *= *this;
        return a;
    }
    // for prime MOD
    mint inv() const {
        return pow(MOD-2);
    }
    mint& operator/=(const mint& a) {
        return (*this) *= a.inv();
    }
    mint operator/(const mint& a) const {
        mint res(*this);
        return res/=a;
    }

    friend ostream& operator<<(ostream& os, const mint& m){
        os << m.x;
        return os;
    }
};



// 階乗を計算する
vector<mint> _factionals = {0,1};
mint factional(long long n) {
    if(_factionals.size() <= n) {
        long long  i = _factionals.size();
        _factionals.resize(n + 1);
        for(; i <= n; i++) {
            _factionals[i] = _factionals[i - 1] * i;
        }
    }
    return _factionals[n];
}

// 組み合わせ数を計算する。nCr
map<long long, vector<mint>> _combinations;
mint combination(long long n, long long r) {
    if(n < 1 || r < 1 || n < r) return mint(0);
    if(n == r) return mint(1);
    
    // nが小さい場合は公式を使う
    if(n < 51000) return factional(n) / (factional(r) * factional(n - r)); 

    // nが大きい場合はnC(r) = nC(r - 1) 
    if(r > n - r) r = n - r;
    if(!_combinations.count(n)) {
        vector<mint> vec = {0, n};
        _combinations.emplace_hint(_combinations.end(), n, vec);
    }
    auto&& vec = _combinations.find(n)->second;
    if(vec.size() <= r) {
        long long i = vec.size();
        vec.resize(r + 1);
        for(; i <= r; i++) {
            vec[i] = (vec[i - 1] * (n - i + 1)) / i;
        }
    }

    return vec[r];
}

// 順序のパターン数を計算する。nPr
map<long long, vector<mint>> _permutations;
mint permutation(long long n, long long r) {
    if(n < 1 || r < 1 || n < r) return mint(0);

    // nが小さい場合は公式を使う
    if(n < 51000) return factional(n) / factional(n - r); 
    if(!_permutations.count(n)) {
        vector<mint> vec = {0, n};
        _permutations.emplace_hint(_permutations.end(), n, vec);
    }
    auto&& vec = _permutations.find(n)->second;
    if(vec.size() <= r) {
        long long i = vec.size();
        vec.resize(r + 1);
        for(; i <= r; i++) {
            vec[i] = vec[i - 1] * (n - i + 1);
        }
    }

    return vec[r];
}


int n,a,b;
mint result;
int main() {
    cin.tie(0); ios::sync_with_stdio(false); cout << fixed << std::setprecision(20);

    cin >> n >> a >> b;
    cout << mint(2).pow(n) - 1 - combination(n, a) - combination(n, b);
}