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)

#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;

#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define ALL(V) (V).begin(),(V).end()
#define SORT(V) sort(ALL(V)) //小さい方からソート
#define REV(V) reverse(ALL(V)) //リバース
#define RSORT(V) SORT(V);REV(V) //大きい方からソート
#define NPN(V) next_permutation(ALL(V)) //順列
#define pb(n) push_back(n)
#define endl '\n'
#define Endl '\n'
#define DUMP(x)  cout << #x << " = " << (x) << endl
#define YES(n) cout << ((n) ? "YES" : "NO"  ) << endl
#define Yes(n) cout << ((n) ? "Yes" : "No"  ) << endl
#define yes(n) cout << ((n) ? "yes" : "no"  ) << endl
#define Yay(n) cout << ((n) ? "Yay!": ":(") << endl
#define RAPID cin.tie(0);ios::sync_with_stdio(false)
#define VSUM(V) accumulate(ALL(V), 0)
#define MID(a,b,c) (a) < (b) && (b) < (c)
#define MIDe(a,b,c) (a) <= (b) && (b) <= (c)
#define IN(n) cin >> n
#define IN2(a,b) cin >> a >> b
#define IN3(a,b,c) cin >> a >> b >> c 
#define IN4(a,b,c,d) cin >> a >> b >> c >> d
#define VIN(V) for(int i = 0; i < (V).size(); i++) {cin >> (V).at(i);}
#define OUT(n) cout << n << endl
#define VOUT(V) REP(i, (V).size()) {cout << (V)[i] << endl;}
#define VOUT2(V) REP(i, (V).size()) {cout << (V)[i] << " ";} cout<<endl;
#define FIX cout << fixed << setprecision(15)

// 型マクロ定義
#define int long long
#define P pair<ll,ll>
#define Vi vector<ll>
#define VVi vector<vector<ll>>
#define Vd vector<double>
#define Vb vector<bool>
#define Vs vector<string>
#define Vc vector<char>
#define M map<ll,ll>
#define S set<ll>
#define PQ priority_queue<ll>
#define PQG priority_queue<ll,V,greater<ll>
//

template<class T> inline bool chmin(T& a, T b) {if(a>b){a=b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b) {if(a<b){a=b;return true;}return false;}

using ll = long long;
using Graph = vector<vector<int>>;

const int MOD = 1000000007;
const int fMOD = 1000000007;
const int INF = 1061109567;
const double EPS = 1e-10;
const double PI  = acos(-1.0);
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

int gcd(int a,int b){return b != 0 ? gcd(b, a % b) : a;} //最大公約数
int lcm(int a, int b){return a * b / gcd(a, b);}
string toStrUp(string str){char diff = 'A'-'a';REP(i,str.size()) str[i] += diff;return str;}
string gCh(string str, int key){return str.substr(key,1);}

int exp_mod(long long a, long long b, int m) {
	long long res = 1;
	while (b > 0) {
		if (b&1)
			res = (res * a)%m;
		a = (a * a)%m;
		b >>= 1;
	}
	return (int)res;
}

int fact_mod(int n, int m) {
	long long res = 1;
	for (int i=n; i>0; --i)
		res = (res * i)%m;
	return (int)res;
}

int comb_mod(int n, int r) {
	long long res = 1;
	for (int i=0; i<r; ++i)
		res = (res * (n - i) )%MOD;
	return ( res * exp_mod(fact_mod(r, MOD), MOD-2, MOD) ) % MOD;
}

signed main() {
  RAPID;
  // デフォルト変数定義
  int n=0,m=0,a=0,b=0,c=0,d=0,x=1,y=0,z=0;
  string s="",t="";
  //

  // ここから
  IN3(n,a,b);
  REP(i,n){
    x *= 2;
    x %= MOD;
  }
  int ans = x-comb_mod(n,a)-comb_mod(n,b)-1;
  if(ans >= 0) OUT(ans%MOD);
  else OUT((ans+MOD)%MOD);
}