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 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#define ll long long
#define rep(i,a,n) for(i=a;i<n;i++)
#define rrep(i,a,n) for(i=a;i>=n;i--)
ll mod = 1000000007;
ll f[101010101], rf[101010101];
ll co[101010101];
ll inv(ll x) {
    ll res = 1;
    ll k = mod - 2;
    ll y = x;
    while (k) {
        if (k & 1) res = (res * y) % mod;
        y = (y * y) % mod;
        k /= 2;
    }
    return res;
}
void init(ll n) {
    f[0] = 1;
    ll i;
    rep(i, 1, n+1) f[i] = (f[i - 1] * i) % mod;
    rf[n] = inv(f[n+1 - 1]);
    rrep(i, n-1, 0) rf[i] = (rf[i + 1] * (i + 1)) % mod;
}
ll C(ll n, ll k) {
    ll a = f[n]%mod; // = n!
    ll b = rf[n-k]%mod; // = (n-k)!
    ll c = rf[k]%mod; // = k!

    ll bc = (b * c) % mod;

    return (a * bc) % mod;
}
int main(){
    ll a,b,n,i,j,ans=1;
    scanf("%lld%lld%lld",&n,&a,&b);
    init(n);
    for(i=0;i<n;i++){
        ans*=2;
        ans%=mod;
    }
    ans=ans-1-C(n,a)-C(n,b);
    printf("%lld\n",ans);
    return 0;
}