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: Accepted)

#include <bits/stdc++.h>
#define rep3(i, s, n, a) for (int i = (s); i < (int)(n); i += a)
#define rep2(i, s, n) rep3(i, s, n, 1)
#define rep(i, n) rep2(i, 0, n)

#define MOD 1e9+7

using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int, int>;

ll factorial(ll n);
void rfactorial(ll n);
ll binary_square(ll x, ll n);

vector<ll> fact(2e5+1, -1), rfact;

int main() {
    ll n , a ,b;
    cin >> n>> a >> b;
    ll x = factorial(b+1);
    rfactorial(b+1);
    ll ans=lldiv(binary_square(2, n) - 1, MOD).rem;
    //cout << ans << endl;
    ll chi=1;
    rep2(i, 1, a+1){
        chi = lldiv(chi * (n - i + 1), MOD).rem;
        //par = lldiv(par * i, MOD).rem;
        // if(i!=a&i!=b){
        //     ans += lldiv(lldiv(x * rfact[i], MOD).rem * rfact[n - i], MOD).rem;
        // }
    }
    // cout << ans << " " << lldiv(chi * rfact[a], MOD).rem  << endl;
    ans = lldiv(ans - lldiv(chi * rfact[a], MOD).rem, MOD).rem;
    if(ans<0){
        ans += MOD;
    }
    // cout << ans << endl;
    rep2(i, a+1, b+1){
        chi = lldiv(chi * (n - i + 1), MOD).rem;
    }
    ans = lldiv(ans - lldiv(chi * rfact[b], MOD).rem, MOD).rem;
    if (ans < 0) {
      ans += MOD;
    }
    cout << ans << endl;
    return 0;
}

ll factorial(ll n) {
  if (n == 0) {
    return 1;
  } else {
    if (fact[n - 1] == -1) {
      fact[n - 1] = factorial(n - 1);
    }
    fact[n] = lldiv(n * fact[n - 1], MOD).rem;
    return fact[n];
  }
}

void rfactorial(ll n) {
  rep(i, n) { rfact.push_back(binary_square(fact[i], MOD - 2)); }
}

ll binary_square(ll x, ll n) {
  ll m = MOD;
  ll ans = 1;
  while (n) {
    if (n % 2 == 1) {
      ans = ans * x % m;
    }
    x = x * x % m;
    n >>= 1;
  }
  return ans % m;
}