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 fup(i,a,b) for(int i=a;i<=b;++i)
#define fdw(i,a,b) for(int i=a;i>=b;--i)
#define sync ios::sync_with_stdio(false);cin.tie(0)
#define cl(a) memset(a,0,sizeof(a))
#define gcd __gcd
#define fi first
#define se second
#define pb push_back
#pragma GCC optimize(2)
using namespace std;
typedef long long ll;
//ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll ksm(ll a,ll b,ll mod){ll ans=1;a%=mod;for(;b;b>>=1){if(b&1) ans=ans*a%mod;a=a*a%mod;}return ans;}
//ll oula(ll n){ ll res=n;for(ll i=2;i*i<=n;i++){if(n%i==0){ res=res-res/i;while(n%i==0) n/=i;}}if(n>1) res=res-res/n;return res;}
//bool MRT(ll x){if(x==2)return true;fup(i,1,50){ll now=rand()%(x-2)+2;if(ksm(now,x-1,x)!=1)return false;}return true;}
//void shai(int n)fup(i,2,n){if(isprime[i]){v.push_back(i);for(int j=2*i;j<=10000;j+=i)isprime[j]=0;}}//"埃氏筛筛选素数"
const int mod=1e9+7;
const int inf=0x3f3f3f3f;
const double eps=1e-7;
const int N=3e5+10;
int a[110];
ll plzh(ll n,ll m)
{
    ll ans=1;
    ll ls=m;
    while(ls)
    {
        ans=(ans%mod*n%mod)%mod;
        n--;
        ls--;
    }
    return ans;
}
int main()
{
//    freopen("D:\\untitled3\\in","r",stdin);
    sync;
    ll n,a,b;
    cin>>n>>a>>b;
    ll ans=0,ls=0;
    ans=(ksm(2,n,mod)-1+mod)%mod;
    ls=(plzh(n,a)%mod*ksm(plzh(a,a),mod-2,mod)%mod)%mod;
    ans=(ans-ls+mod)%mod;
    ls=(plzh(n,b)%mod*ksm(plzh(b,b),mod-2,mod)%mod)%mod;
    ans=(ans-ls+mod)%mod;
    cout<<ans<<endl;
    return 0;
}