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: Wrong Answer)

#include <stdio.h>

#define LIMIT 1000000007

/* n! */
long factorial(long n){
    long x = 1;
    for(long i=1; i<=n; i++){
        x *= i;
        x %= LIMIT;
    }
    return x;
}

long nPk(long n, long k){
    long x=1, i;
    for(i=0; i<k; i++){
        x *= (n-i);
        x %= LIMIT;
    }
    return x;
}

long modpow(long a, long n, long p){
    if(n==1) return a % p;
    if(n%2 == 1) return (a * modpow(a, n-1, p)) % p;

    long t = modpow(a, n/2, p);
    return (t*t) % p;
}

/*

nCk mod(LIMIT) = nPk/k! mod(LIMIT) = X/Y mod(LIMIT)
Y * Y^LIMIT-2 mod(LIMIT) = 1 フェルマーの小定理
>> nCk mod(LIMIT) = X * Y^LIMIT-2 mod(LIMIT)

*/

int main(){
    long n, a, b;
    long ans = 0;
    scanf("%ld%ld%ld", &n, &a, &b);

    ans += modpow(2, n, LIMIT) - 1;
    long x, y;

    x = nPk(n, a);
    y = factorial(a);
    ans -= (x * modpow(y, LIMIT-2, LIMIT))%LIMIT;

    x = nPk(n, b);
    y = factorial(b);
    ans -= (x * modpow(y, LIMIT-2, LIMIT))%LIMIT;

    if(ans<0) ans += LIMIT;

    printf("%ld\n", ans%LIMIT);
    return 0;
}