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>
#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define rrep(i,x) for(int i=((int)(x)-1);i>=0;i--)
//#define MAX 200001
using namespace std;
using ll = long long;

const int MOD = 1000000007;
const int MAX = 9999999;
const ll INF = numeric_limits<ll>::max();
const int inf = 1e8;

//a^nをpで割ったあまり
ll modPow(ll a, ll n, ll p) {
    ll ans = 1;
    while(n != 0){
        if(n&1) ans = ans*a % p;
        a = a*a % p;
        n = n >> 1;
    } 
    return ans;
}

//nCrをMODで割った余り
ll nCr(ll n, ll r,ll mod){
    ll ans=0,x=1,y=1;
    for(ll i=1;i<=r;i++){
        x=x*(n-i+1)%mod;
        y=y*i%mod;
    }
    ans=x*modPow(y,mod-2,mod);
    return ans;
}

int main(){
    cin.tie( 0 ); ios::sync_with_stdio( false );

    ll n,a,b;
    cin >> n >> a >> b;
    ll ans=0;
    ans=modPow(2,n,MOD)-nCr(n,a,MOD)-nCr(n,b,MOD)-1;
    while (ans < 0) ans += MOD;
    cout << ans;
}