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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.lang.Math;
import java.util.Scanner;
class Main {
	
    public static void main(String[] args) {
    	Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a = sc.nextInt();
        int b = sc.nextInt();
        long sum = 1;
        if (n%2 != 0) {
        		for (int i=1; i<=n/2; i++) {
        			int num = 1;
        			for (int j=1; j<=i; j++) {
        				num = num * (n-j+1) / j;
        			}
        			sum += num*2;
        		}
        		int aa = 1;
        		for (int j=1; j<=a; j++) {
    				aa = aa * (n-j+1) / j;
    			}
        		int bb = 1;
        		for (int j=1; j<=b; j++) {
    				bb = bb * (n-j+1) / j;
    			}
        		sum = sum - aa - bb;
        		System.out.println(sum%((int)Math.pow(10, 9)+7));
        } else {
        	for (int i=1; i<=n/2; i++) {
    			int num = 1;
    			for (int j=1; j<=i; j++) {
    			
    				num = num * (n-j+1) / j;
    			}
    			if (i == n/2) {
    				sum += num;
    			} else {
    				sum += num*2;
    			}
    			
    		}
    		int aa = 1;
    		for (int j=1; j<=a; j++) {
				aa = aa * (n-j+1) / j;
			}
    		int bb = 1;
    		for (int j=1; j<=b; j++) {
				bb = bb * (n-j+1) / j;
			}
    		sum = sum - aa - bb;
    		System.out.println(sum%((int)Math.pow(10, 9)+7));
        }
    }
}