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++ (GCC 9.2.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")

using namespace std;
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
typedef long long ll;
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define REP(i,num,n) for(int i=num, i##_len=(n); i<i##_len; ++i)
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define print(x) (cout << (x) << endl)
#define pb push_back
#define mp make_pair
#define sz(x) int(x.size())
template<class T> inline void Yes(T condition){ if(condition) cout << "Yes" << endl; else cout << "No" << endl; }
#define fs first
#define sc second
// 多倍長
#include <boost/rational.hpp>
#include <boost/multiprecision/cpp_int.hpp>
using boost::rational;
using boost::multiprecision::cpp_int;

const int INF = 1e9;
const ll LLINF = 1e16;
const ll MOD = 1e9+7;
template<class T> inline void add(T &a, T b){a = ((a+b) % MOD + MOD) % MOD;};

template<typename T>
vector<T> make_v(size_t a){return vector<T>(a);}
 
template<typename T,typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v<T>(ts...))>(a,make_v<T>(ts...));
}
 

// 指定した文字cが文字列に何文字入ってるか
ll stringcount(string s, char c) {
    return count(s.cbegin(), s.cend(), c);
}


// 割り算の切り上げ処理
template<typename T>
T rp(T a, T b) {
  return (a + b - 1) / b;
}

// 桁和
template<typename T>
T digsum(T n) {
    T res = 0;
    while(n > 0) {
        res += n%10;
        n /= 10;
    }
    return res;
}

// 回文判定
bool kai (string s) {
    string t{s};
    reverse(begin(t), end(t));
    return s == t;
}

using pint = pair<int, int>;
using pll = pair<ll, ll>;


template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator < (const Fp& x) const {
        return this->val < x.val;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};

// 二項係数ライブラリ
template<class T> struct BiCoef {
    vector<T> fact_, inv_, finv_;
    constexpr BiCoef() {}
    constexpr BiCoef(int n) noexcept : fact_(n, 1), inv_(n, 1), finv_(n, 1) {
        init(n);
    }
    constexpr void init(int n) noexcept {
        fact_.assign(n, 1), inv_.assign(n, 1), finv_.assign(n, 1);
        int MOD = fact_[0].getmod();
        for(int i = 2; i < n; i++){
            fact_[i] = fact_[i-1] * i;
            inv_[i] = -inv_[MOD%i] * (MOD/i);
            finv_[i] = finv_[i-1] * inv_[i];
        }
    }
    constexpr T com(int n, int k) const noexcept {
        if (n < k || n < 0 || k < 0) return 0;
        return fact_[n] * finv_[k] * finv_[n-k];
    }
    constexpr T fact(int n) const noexcept {
        if (n < 0) return 0;
        return fact_[n];
    }
    constexpr T inv(int n) const noexcept {
        if (n < 0) return 0;
        return inv_[n];
    }
    constexpr T finv(int n) const noexcept {
        if (n < 0) return 0;
        return finv_[n];
    }
};

using mint = Fp<MOD>;
BiCoef<mint> bc;



int main(void)
{
    cout.tie(0);
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(20);
    ll n,a,b;
    cin >> n >> a >> b;
    bc.init(n + 1);
    mint ans = 1;
    rep(i, n){
        ans *= 2;
    }
    ans -= 1;
    ans -= bc.com(n, a);
    ans -= bc.com(n, b);
    print(ans);

}