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++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl 	"\n"

typedef vector< ll > 	vi;
typedef vector< vi > 	vvi;
typedef pair< ll,ll > 	ii;
#define pb 		push_back
#define all(c) 		(c).begin(),(c).end()
#define tr(c,i) 	for(typeof((c).begin() i = (c).begin(); i != (c).end(); i++)
#define present(c,x) 	((c).find(x) != (c).end())

#define MOD 	1000000007
#define INF 	(int)1e9
#define MEM(a, b) memset(a, (b), sizeof(a))

ll binomialCoeffSum(ll n) 
{ 
    return (1 << n); 
} 
ll nCrModpDP(ll n, ll r, ll p) 
{ 
    // The array C is going to store last row of 
    // pascal triangle at the end. And last entry 
    // of last row is nCr 
    ll C[r+1]; 
    memset(C, 0, sizeof(C)); 
  
    C[0] = 1; // Top row of Pascal Triangle 
  
    // One by constructs remaining rows of Pascal 
    // Triangle from top to bottom 
    for (ll i = 1; i <= n; i++) 
    { 
        // Fill entries of current row using previous 
        // row values 
        for (ll j = min(i, r); j > 0; j--) 
  
            // nCj = (n-1)Cj + (n-1)C(j-1); 
            C[j] = (C[j] + C[j-1])%p; 
    } 
    return C[r]; 
} 
  
// Lucas Theorem based function that returns nCr % p 
// This function works like decimal to binary conversion 
// recursive function.  First we compute last digits of 
// n and r in base p, then recur for remaining digits 
ll nCrModpLucas(ll n, ll r, ll p) 
{ 
   // Base case 
   if (r==0) 
      return 1; 
  
   // Compute last digits of n and r in base p 
   ll ni = n%p, ri = r%p; 
  
   // Compute result for last digits computed above, and 
   // for remaining digits.  Multiply the two results and 
   // compute the result of multiplication in modulo p. 
   return (nCrModpLucas(n/p, r/p, p) * // Last digits of n and r 
           nCrModpDP(ni, ri, p)) % p;  // Remaining digits 
} 
int main()
{	 ios_base::sync_with_stdio(false);
	 cin.tie(NULL);
	
	ll n,a,b;
	cin>>n>>a>>b;

	ll ans= binomialCoeffSum(n)%MOD;
	ans -= nCrModpLucas(n,0,MOD);
	ans -= nCrModpLucas(n,a,MOD);
	ans -= nCrModpLucas(n,b,MOD);
	
	if(ans<=0)
		cout<<"0";
	else cout<<ans;

	return 0;
}