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 <vector>
#include <iostream>
#include <cmath>
#include <map>
#include <unordered_map>
#include <algorithm>
#include <fstream>
#include <unistd.h>
#include <string.h>
#include <string>
#include <numeric>
#include <queue>
#include <deque>
#include <sstream>
#include <iomanip>
#include <set>
#include <stack>
#include <cassert>
#include <functional>
#include <random>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<vvl> vvvl;
typedef vector<vvvl> vvvvl;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<bool> vb;
typedef vector<vector<bool>> vvb;
typedef vector<vvb> vvvb;
typedef vector<vvvb> vvvvb;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef vector<vvs> vvvs;
typedef vector<char> vc;
typedef vector<vector<char>> vvc;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<vvd> vvvd;
typedef long double LD;
typedef vector<LD> vld;
typedef vector<pair<LD, LD>> vpld;
typedef pair<ll, ll> PL;
typedef vector<PL> vpl;
typedef vector<vector<PL>> vvpl;
typedef vector<vvpl> vvvpl;
typedef pair<ll, PL> TL;
typedef vector<TL> vtl;
typedef vector<vtl> vvtl;
typedef long double ld;
typedef vector<ld> vld;
const ll LINF = 1e17;
const double pi = 3.1415926535897932;
const string endstr = "\n";
#define FOR(i, a, b) for(ll i = (a); i < b; i++)
#define RFOR(i, a, b) for(ll i = (a); i > b; i--)
#define rep(i, n) for(ll i = 0; i < n; i++)
#define rrep(i, n) for(ll i = n-1; i > -1; i--)
#define FORMAP(it, m) for(auto it = m.begin(); it != m.end(); it++)
#define ff first
#define ss second
#define pb push_back
#define epb emplace_back
#define ALL(X) (X).begin(),(X).end()

template <typename T> T gcd(T a, T b) {return (a == 0) ? b : gcd(b%a, a);}
template <typename T> T lcm(T a, T b) { return a / gcd(a, b) * b;}

bool p_comp_fs(const PL p1, const PL p2){ return p1.first < p2.first;};
bool p_comp_fg(const PL p1, const PL p2){ return p1.first > p2.first;};
bool p_comp_ss(const PL p1, const PL p2){ return p1.second < p2.second;};
bool p_comp_sg(const PL p1, const PL p2){ return p1.second > p2.second;};
template <typename T>
vector<T> uniquen(vector<T> vec /* copy */){
    sort(ALL(vec)); vec.erase(unique(vec.begin(), vec.end()), vec.end()); return vec;
}

inline ll popcnt(ll x){return __builtin_popcountll((unsigned long long)x);};
template<class T> bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }
//friend bool operator<(const tpl& lhs, const tpl& rhs){
//    return std::tie(lhs.l, lhs.r) < std::tie(rhs.l, rhs.r);
//}

bool bit(ll st, ll b){return ((st>>b)&1) == 1;}


const ll mod = 1e9+7;
template< int mod >
struct ModInt {
    int x;
    ModInt() : x(0) {}
    ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
    ModInt &operator+=(const ModInt &p) {
        if((x += p.x) >= mod) x -= mod;
        return *this;
    }
    ModInt &operator-=(const ModInt &p) {
        if((x += mod - p.x) >= mod) x -= mod;
        return *this;
    }
    ModInt &operator*=(const ModInt &p) {
        x = (int) (1LL * x * p.x % mod);
        return *this;
    }
    ModInt &operator/=(const ModInt &p) {
        *this *= p.inv();
        return *this;
    }
    ModInt operator-() const { return ModInt(-x); }
    ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
    ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
    ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
    ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
    bool operator==(const ModInt &p) const { return x == p.x; }
    bool operator!=(const ModInt &p) const { return x != p.x; }
    
    ModInt inv() const {
        int a = x, b = mod, u = 1, v = 0, t;
        while(b > 0) {
            t = a / b;
            swap(a -= t * b, b);
            swap(u -= t * v, v);
        }
        return ModInt(u);
    }
    
    ModInt pow(int64_t n) const {
        ModInt ret(1), mul(x);
        while(n > 0) {
            if(n & 1) ret *= mul;
            mul *= mul;
            n >>= 1;
        }
        return ret;
    }
    
    friend ostream &operator<<(ostream &os, const ModInt &p) {
        return os << p.x;
    }
    
    friend istream &operator>>(istream &is, ModInt &a) {
        int64_t t;
        is >> t;
        a = ModInt< mod >(t);
        return (is);
    }
    
    static int get_mod() { return mod; }
};

using mint = ModInt<mod>;
typedef vector<mint> vm;
typedef vector<vm> vvm;
typedef vector<vvm> vvvm;

const ll size = 300300; // ******** CHECK ME !!!!!!! ********
mint fact[size], finv[size];

mint nCk(ll n, ll k){
    if(n < k) return 0;
    return fact[n]*finv[k]*finv[n-k];
}

mint Catalan(ll N){
    return nCk(2*N, N) - nCk(2*N, N-1);
}

void prepare(){
    fact[0] = 1;
    rep(i, size-1) fact[i+1] = fact[i]*mint(i+1);
    finv[size-1] = fact[size-1].inv();
    for (ll i = size-1; i > 0; i--) {
        finv[i-1] = finv[i]*mint(i);
    }
}

mint nck(ll N, ll K){
    mint res = 1;
    for(ll i = N; i > N-K; i--) res *= i;
    for(ll i = 1; i <= K; i++) res *= mint(i).inv();
    return res;
}

int main(){
#ifdef __DEBUG_LOCAL__
    ifstream file("../../../../../in.txt"); if(file.is_open()) cin.rdbuf(file.rdbuf());
#endif
    ios_base::sync_with_stdio(false); cin.tie(nullptr);
    prepare();
    ll N, A, B; cin >> N >> A >> B;
    
    mint ans = mint(2).pow(N) - 1;
    ans -= nck(N, A);
    ans -= nck(N, B);
    cout << ans << endstr;
    return 0;
}