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: Wrong Answer)

import java.util.*;
public class Main {
        static Scanner sc = new Scanner(System.in);
        static void myout(Object t){System.out.println(t);}//standard output
        static void myerr(Object t){System.err.println(t);}//standard error
        static String getStr(){return sc.next();}
        static int getInt(){return Integer.parseInt(getStr());}
        static long getLong(){return Long.parseLong(getStr());}
        static boolean hasNext(){return sc.hasNext();}
        static char[] mySplit(String str){return str.toCharArray();}
        public static void main(String[] args){
          int N = getInt();
          int A = getInt();
          int B = getInt();
          if(N == 2){
            myout(0);
            return;
          }
          int mod = 1000000007;
          int aResult = originComb(N,A,mod);
          int bResult = originComb(N,B,mod);
          int output = ((int)(N * N + 1)%mod) - aResult - bResult;
          if(output < 0){
            output += mod;
          }
          myout(output);
        }
        //Method addition frame start
static int originComb(int n, int k, int m){
  int[] numerator = new int[k];
  int[] denominator = new int[k];
  for(int i = 0; i < k; i++){
    numerator[i] = n - k + i + 1;
    denominator[i] = i + 1;
  }
  for(int i = 2; i <= k; i++){
    int pivot = denominator[i - 1];
    if(pivot > 1){
      int offset = (n - k) % i;
      for(int j = i - 1; j < k; j += i){
        numerator[j - offset] /= pivot;
        denominator[j] /= pivot;
      }
    }
  }
  int res = 1;
  for(int i = 0; i < k; i++){
    if(numerator[i] > 1){
      res = res * numerator[i] % m;
    }
  }
  return res;
}
  static long originPow(long x, long n) {return originPow(x,n,-1);}
static long originPow(long x, long n,int m) {
        long ans = 1;
        while (n > 0) {
                if ((n & 1) == 1){
                  ans = ans * x;
                  if(m != -1){ans %= m;}
                }
                x = x * x;
                if(m != -1){x %= m;}
                n >>= 1;
        }
        return ans;
}
        //Method addition frame end
}