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 M (1000000007LL)

/***************************************
    繰り返し二乗法
    (a**n%m)
***************************************/
long long powm( long a, long n, long m )
{
    long long ret = 1;
    for( ; n>0; n>>=1, a=(long long)a*a%m ) {
        if( n % 2 == 1 ) {
            ret = ret * a % m;
        }
    }

    return( ret );
}

/***************************************
    aからbまでの掛け算を計算
    ((a(a+1)(a+2)…(b))%m)
***************************************/
long long multiplem( long a, long b, long m )
{
    int i;
    long long ret = 1;

    if( a > b ) {
        i = a;
        a = b;
        b = i;
    }

    for( i=a; i<=b; i++ ) {
        ret = (ret * i) %m;
    }

    return( ret );
}

/***************************************
    組み合わせ計算(nCa%m)
***************************************/
long long combinationm( long n, long a, long m )
{
    int i;
    long long ret;

    if( n-a < a ) a = n-a;

    // 分母を計算
    ret = multiplem( n, n-a+1, m );

    // 1/(a!)を計算して嬰かける
    for( i=1; i<=a; i++ ) {
        ret = (ret * powm( i, m-2, m ) ) % m;
    }

    return( ret );
}

/***************************************
    main
***************************************/
int main( int argc, char *argv[] )
{
    long n, a, b;
    long long ret;

    scanf( "%d%d%d", &n, &a, &b );

    ret = powm( 2, n, M ) - 1;
    ret -= combinationm( n, a, M );
    ret -= combinationm( n, b, M );

    if( ret < 0 ) ret+= M;

    printf( "%lld\n", ret );
}