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)

/*Function Template*/
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef pair<ll,ll> pint;
const int MAX = 510000;
const int MOD = 1000000007;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define Rep(i, n) for(ll i = 1; i < (n); i++)
#define ALL(a)  (a).begin(),(a).end()
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define PI 3.14159265358979323846
ll fac[MAX], finv[MAX], inv[MAX];


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;
}
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }

}
ll Len(ll n) {
  ll s=0;
  while(n!=0) s++, n/=10;
  return s;
}


ll Sint(ll n) {
  ll m=0,s=0,a=n;
  while(a!=0) s++, a/=10;
  for(ll i=s-1;i>=0;i--) m+=n/((ll)pow(10,i))-(n/((ll)pow(10,i+1)))*10;
  return m;
}


ll Svec(vector<ll> v){
  ll n=0;
  for(ll i=0;i<v.size();i++) n+=v[i];
  return n;
}


ll GCD(ll a,ll b) {
  return b ? GCD(b,a%b) : a;
}


ll LCM(ll a,ll b){
  return a/GCD(a,b)*b;
}


ll Factorial(ll n){
  ll m=1;
  while(n>=1) m*=n,n--;
  return m;
}

void runlength(string s,vector<pair<char,ll>> &p){
  ll x=1;
  if(s.size()==1){
    p.push_back(pair<char,ll>(s[0],1));
  }
  for(ll i=0;i<s.size()-1;i++){
    if(s[i]==s[i+1]){
      x++;
      if(i==s.size()-2){
        p.push_back(pair<char,ll>(s[i],x));
      }
    }else{
      p.push_back(pair<char,ll>(s[i],x));
      x=1;
      if(i==s.size()-2){
        p.push_back(pair<char,ll>(s[s.size()-1],x));
      }
    }
  }
}

ll COM(ll n,ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}


string Toupper(string s){
  string ans="";
  rep(i,s.size()){
    if('a'<=s[i] && s[i]<='z') ans+=(char)s[i]-32;
    else ans+=s[i];
  }
  return ans;
}
string Tolower(string s){
  string ans="";
  rep(i,s.size()){
    if('A'<=s[i] && s[i]<='Z') ans+=(char)s[i]+32;
    else ans+=s[i];
  }
  return ans;
}


const int MAX_N=100010;
vector<bool> sieve_of_eratosthenes(){
  vector<bool> isPrime(MAX_N+1,true);
  for(int i=2;i<=MAX_N;i++){
    if(isPrime[i]){
      for(int j=2*i;j<=MAX_N;j+=i){
        isPrime[j]=false;
      }
    }
  }
  return isPrime;
}

vector<pint> prime_factorize(ll n){
  vector<pint> ans;
  for(ll p=2;p<=sqrt(n);p++){
    if(n%p!=0) continue;
    ll cnt=0;
    while(n%p==0){
      n/=p;
      cnt++;
    }
    ans.push_back(make_pair(p,cnt));
  }
  if(n!=1) ans.push_back(make_pair(n,1));
  return ans;
}

/*bool cmp(pint a, pint b) { return a.second < b.second; }*/

template<int MOD> struct Fp {
    long long val;
    constexpr Fp(long long v = 0) noexcept : val(v % MOD) {
        if (val < 0) val += MOD;
    }
    constexpr int getmod() { return MOD; }
    constexpr Fp operator - () const noexcept {
        return val ? MOD - val : 0;
    }
    constexpr Fp operator + (const Fp& r) const noexcept { return Fp(*this) += r; }
    constexpr Fp operator - (const Fp& r) const noexcept { return Fp(*this) -= r; }
    constexpr Fp operator * (const Fp& r) const noexcept { return Fp(*this) *= r; }
    constexpr Fp operator / (const Fp& r) const noexcept { return Fp(*this) /= r; }
    constexpr Fp& operator += (const Fp& r) noexcept {
        val += r.val;
        if (val >= MOD) val -= MOD;
        return *this;
    }
    constexpr Fp& operator -= (const Fp& r) noexcept {
        val -= r.val;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr Fp& operator *= (const Fp& r) noexcept {
        val = val * r.val % MOD;
        return *this;
    }
    constexpr Fp& operator /= (const Fp& r) noexcept {
        long long a = r.val, b = MOD, u = 1, v = 0;
        while (b) {
            long long t = a / b;
            a -= t * b; swap(a, b);
            u -= t * v; swap(u, v);
        }
        val = val * u % MOD;
        if (val < 0) val += MOD;
        return *this;
    }
    constexpr bool operator == (const Fp& r) const noexcept {
        return this->val == r.val;
    }
    constexpr bool operator != (const Fp& r) const noexcept {
        return this->val != r.val;
    }
    friend constexpr ostream& operator << (ostream &os, const Fp<MOD>& x) noexcept {
        return os << x.val;
    }
    friend constexpr Fp<MOD> modpow(const Fp<MOD> &a, long long n) noexcept {
        if (n == 0) return 1;
        auto t = modpow(a, n / 2);
        t = t * t;
        if (n & 1) t = t * a;
        return t;
    }
};


using mint = Fp<MOD>;


ll modpow(ll a, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
/*↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓*/
int main() {
  IOS;
  COMinit();
  ll n,a,b;cin>>n>>a>>b;COMinit();ll ans=modpow(2,n,MOD);
  ans--;
  mint ans1=1;
  mint ans2=1;
  rep(i,a) ans1=ans1*(n-i)/(i+1);
  rep(i,b) ans2=ans2*(n-i)/(i+1);
  mint nnn=ans;
  cout<<nnn-ans1-ans2<<endl;
}