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: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < n; i++)
typedef long long ll;
typedef pair<int, int> PI;
typedef pair<ll, ll> PL;

const ll mod = 1000000007;

int n;
ll powmod(int a, int n, ll m){ //a^nのmodを計算する
	ll res = 1;
	while (n > 0){
		if(n & 1) res = res * a % mod;
		a = a * a % mod;
		n >> 1;
	}
	return res;
}

ll facmod(int a, int b, ll m){ //a*(a+1)*...*(b-1)*bのmod
	ll res = 1; int cnt = a;
	if(b >= cnt)res = res * a % mod;
	cnt++;
	return res;
}

ll solve(int k){ // nCk = s/t % mod == k1/k2について逆元を使ってnCkを出す
	int k1, k2;
	k1 = facmod(1, k, mod);
	k2 = facmod(n - k + 1, n, mod);
	ll res = powmod(k2, mod - 2, mod); //k2の逆元
	res = res * k1 % mod;
	return res;
}

int main() {
	cin.tie(0);
   	ios::sync_with_stdio(false);
    cin >> n;
	int a, b; cin >> a >> b;
	ll tmp = powmod(2, n, mod) - 1; //制約なし（何本使ってもいい）時
	ll a1 = solve(a), b1 = solve(b);
	cout << tmp - a1 - b1 << endl;
}