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)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); ++i)
#define all(x) x.begin(),x.end()
#define ln '\n'
const long double PI = acos(-1.0L);
const long long MOD = 1000000007LL;
// const long long MOD = 998244353LL;
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
template<class T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true;} return false; }
template<class T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true;} return false; }
////////////////////////////////////////////////////////////////////////////////////////////////////////

template <std::uint_fast64_t Modulus> class modint {
  using u64 = std::uint_fast64_t;

public:
  u64 a;

  constexpr modint(const u64 x = 0) noexcept : a(x % Modulus) {}
  constexpr u64 &value() noexcept { return a; }
  constexpr const u64 &value() const noexcept { return a; }
  constexpr modint operator+(const modint rhs) const noexcept {
    return modint(*this) += rhs;
  }
  constexpr modint operator-(const modint rhs) const noexcept {
    return modint(*this) -= rhs;
  }
  constexpr modint operator*(const modint rhs) const noexcept {
    return modint(*this) *= rhs;
  }
  constexpr modint operator/(const modint rhs) const noexcept {
    return modint(*this) /= rhs;
  }
  constexpr modint &operator+=(const modint rhs) noexcept {
    a += rhs.a;
    if (a >= Modulus) {
      a -= Modulus;
    }
    return *this;
  }
  constexpr modint &operator-=(const modint rhs) noexcept {
    if (a < rhs.a) {
      a += Modulus;
    }
    a -= rhs.a;
    return *this;
  }
  constexpr modint &operator*=(const modint rhs) noexcept {
    a = a * rhs.a % Modulus;
    return *this;
  }
  constexpr modint &operator/=(modint rhs) noexcept {
    u64 exp = Modulus - 2;
    while (exp) {
      if (exp % 2) {
        *this *= rhs;
      }
      rhs *= rhs;
      exp /= 2;
    }
    return *this;
  }
};

using mint = modint<MOD>;

int main() {
    ios::sync_with_stdio(false); cin.tie(nullptr);
    int N,A,B; cin >> N >> A >> B;
    int K = N;
    mint ans = 1;
    mint t = 2;
    while (K > 0) {
        if (K&1) ans *= t;
        K /= 2;
        t *= t;
    }

    mint p = 1;
    for (int i = N; i > N-A; i--) {
        p *= i;
    }
    for (int i = 1; i <= A; i++) {
        p /= i;
    }
    mint q = 1;
    for (int i = N; i > N-B; i--) {
        q *= i;
    }
    for (int i = 1; i <= B; i++) {
        q /= i;
    }
    ans -= p+q+1;

    cout << ans.a << ln;
}