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 PI 3.14159265358979323846
#define MAXINF (1e18L)
#define INF (1e9L)
#define EPS (1e-9)
#define MOD ((ll)(1e9+7))
#define REP(i, n) for(int i=0;i<int(n);++i)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define RREP(i, n) for(int i=int(n)-1;i>=0;--i)
#define ALL(v) v.begin(),v.end()
#define FIND(v,x) (binary_search(ALL(v),(x)))
#define SORT(v) sort(ALL(v))
#define RSORT(v) sort(ALL(v));reverse(ALL(v))
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl
#define Yes(n) cout<<((n)?"Yes":"No")<<endl
#define YES(n) cout<<((n)?"YES":"NO")<<endl
#define pb push_back
#define fi first
#define se second
using namespace std;
template<class A>void pr(A a){cout << (a) << endl;}
template<class A,class B>void pr(A a,B b){cout << a << " "  ;pr(b);}
template<class A,class B,class C>void pr(A a,B b,C c){cout << a << " " ;pr(b,c);}
template<class A,class B,class C,class D>void pr(A a,B b,C c,D d){cout << a << " " ;pr(b,c,d);}
template<class T> inline bool chmin(T& a, T b){return a>b ? a=b, true : false;}
template<class T> inline bool chmax(T& a, T b){return a<b ? a=b, true : false;}
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll,ll> pll;

template<uint32_t mod>
struct ModInt{
    uint32_t a;
    ModInt& s(uint32_t vv){
        a=vv<mod?vv:vv-mod;
        return *this;
    }
 
    ModInt(int64_t x=0){s(x%mod+mod);}
 
    ModInt& operator+=(const ModInt &x){return s(a+x.a);}
    ModInt& operator-=(const ModInt &x){return s(a+mod-x.a);}
    ModInt& operator*=(const ModInt &x){
        a=uint64_t(a)*x.a%mod;
        return *this;
    }
    ModInt& operator/=(const ModInt &x){
        *this*=x.inv();
        return *this;
    }
 
    ModInt operator+(const ModInt &x)const{return ModInt(*this)+=x;}
    ModInt operator-(const ModInt &x)const{return ModInt(*this)-=x;}
    ModInt operator*(const ModInt &x)const{return ModInt(*this)*=x;}
    ModInt operator/(const ModInt &x)const{return ModInt(*this)/=x;}
    bool operator==(const ModInt &x)const{return a==x.a;}
    bool operator!=(const ModInt &x)const{return a!=x.a;}
    bool operator<(const ModInt &x)const{return a<x.a;}
 
    ModInt operator-()const{return ModInt()-*this;}
    ModInt pow(int64_t n)const{
        ModInt res(1),x(*this);
        while(n){
            if(n&1)res*=x;
            x*=x;
            n>>=1;
        }
        return res;
    }
 
    ModInt inv()const{return pow(mod-2);}
};
 
template<uint32_t mod>
istream& operator>>(istream& in,ModInt<mod>& a){
    return (in>>a.a);
}
template<uint32_t mod>
ostream& operator<<(ostream& out,const ModInt<mod>& a){
    return (out<<a.a);
}
using mint=ModInt<MOD>;

vector<mint> inv, fact, invfact;
void mod_build(int n = 101010){
    fact.assign(n+1, mint(1));
    inv.assign(n+1, mint(1));
    invfact.assign(n+1, mint(1));
    REP(i, n){
        fact[i+1] = fact[i] * mint(i+1);
        if(i > 0) inv[i+1] = mint(MOD) - inv[MOD % (i+1)] * mint(MOD / (i+1));
        invfact[i+1] = invfact[i] * inv[i+1];
    }
}
mint perm(int n, int k){
    if(n < 0 || k < 0 || k > n) return 0;
    return fact[n] * invfact[n-k];
}
mint comb(int n, int k){
    if(n<0 || k<0 || k>n) return 0;
    return (fact[n] * invfact[n-k]) * invfact[k];
}
mint comb2(ll n, ll k){
    mint X(n), Y(1);
    REP(i, k-1) {
        X *= n-i-1;
        Y *= i+2;
    }
    return X*Y.pow(MOD-2); 
    // mint ret(1);
    // REP(i, k){
    //     ret *= (n-i);
    //     ret *= mint(i+1).pow(MOD - 2);
    // }
    // return ret;
}

int main(void)
{
    ll n,a,b;
    cin >> n >> a >> b;
    mod_build();
    mint m = mint(2).pow(n);
    pr(m-1-comb2(n, a)-comb2(n, b));
}