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: Wrong Answer)

#include<bits/stdc++.h>
#define rep(i,n)for(int i=0;i<n;++i)
#include<string>
#include<vector>
using namespace std;
typedef long long ll;
typedef pair<int, int>P;
//max=({});
//条件式が真ならwhileの中身を回し続ける
//printf("%d\n", ans);

//pairの入力
//vector<pair<ll, ll>>work(n);
//rep(i, n) {
//	ll a, b;
//	cin >> a >> b;
//	work[i] = make_pair(a, b);
//}

int g[15][15];
const int INF = 1001001001;
const int dx[4] = { -1,0,1,0 };
const int dy[4] = { 0,-1,0,1 };

//最大公約数
ll gcd(ll x, ll y) {
	return y ? gcd(y, x % y) : x;
}

//素因数分解
vector<pair<ll, int>>factorize(ll n) {
	vector<pair<ll, int>>res;
	for (ll i = 2;i * i <= n;++i) {
		if (n % i)continue;
		res.emplace_back(i, 0);
		while (n % i == 0) {
			n /= i;
			res.back().second++;
		}
	}
	if (n != 1)res.emplace_back(n, 1);
	return res;
}

int dp[1010][1010];
int bingo[3][3];
bool flag[3][3];

ll n;
ll ans = 0;

void dfs(ll x, int a, int b, int c) {
	if (x > n)return;
	if (a && b && c)ans++;
	dfs(10 * x + 3, 1, b, c);
	dfs(10 * x + 5, a, 1, c);
	dfs(10 * x + 7, a, b, 1);
}

ll comb(ll n, ll x) {
	ll res = 1;
	ll big = max(x, n - x);
	ll little = min(x, n - x);
	for (ll i = n;i >= big + 1;--i) {
		res *= i;
	}
	for (ll i = 1;i <= little;++i) {
		res /= i;
	}
	return res;
}

int main() {
	ll n, a, b;
	cin >> n >> a >> b;
	ll ans = 2;
	ll kazu = n;
	if (n % 2 == 0) {
		while (kazu > 1) {
			ans *= ans;
			kazu /= 2;
		}
		ans -= comb(n, a);
		ans -= comb(n, b);
	}
	else {
		kazu--;
		while (kazu > 1) {
			ans *= ans;
			kazu /= 2;
		}
		ans *= 2;
		ans -= comb(n, a);
		ans -= comb(n, b);
	}
	cout << ans << endl;
	return 0;
}