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 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <stdio.h>

using namespace std;
#define inputInt(a) int a; cin >> a; 
#define inputInt2(a,b) int a; int b; cin >> a >> b; 
#define inputInt3(a,b,c) int a; int b; int c; cin >> a >> b >> c;
#define inputLong(a) long a;cin >> a; 
#define inputIntArray(a,N) int a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputLongArray(a,N) long a[N];for(int i=0;i<N;i++){cin >> a[i];}
#define inputIntArray2(a,b,N) int a[N]; int b[N]; for(int i=0;i<N;i++){cin >> a[i] >> b[i];}
#define output(answer) cout << answer << endl;
#define fN(i,N)    for(int i=0; i<N; i++)
#define fSE(i,s,e) for(int i=s; i<=e; i++)
#define ll long long
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define FOR(i,N) for(int i=0; i<N; i++)
#define cinv(v,N) vector<int> v(N); for(int i=0; i<N; i++){ cin >> v[i]; }
#define mt make_tuple
#define vll vector<long long int>
#define vvll vector<vector<long long int>>
#define all(v) (v).begin(),(v).end()
#define show(v) FOR(i,v.size()){ cout << v[i] << " "; } cout << endl;
#define showshow(v) FOR(i,v.size()){ FOR(j,v[i].size()){ cout << v[i][j] << " "; } cout << endl; }
const ll MOD = 1e9+7;
const ll MOD2 = 998244353;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

//Combination計算 //ComInitしないと何もできません。
const int LIMIT = 1e6;//1e6→50ms,1e7→500ms,1e8→5000ms程度。
//const ll MOD = 1e9+7;
vector<ll> FacMod(LIMIT), InvMod(LIMIT), FacInvMod(LIMIT);
const int dpComTableLIMIT = 3e3+1;
vector<vector<ll>> Com(dpComTableLIMIT,vector<ll>(dpComTableLIMIT,0));
const int ComKTableLIMIT = 1e7;
vector<ll> ComK(ComKTableLIMIT);

void DPComInit(){
  //Com[0][j]は0で初期化されている。
  for(int i=0; i<dpComTableLIMIT; i++){ Com[i][0] = 1; }
  
  for(int i=1; i<dpComTableLIMIT; i++){
    for(int j=1; j<dpComTableLIMIT; j++){
      Com[i][j] = (Com[i-1][j-1]+Com[i-1][j])%MOD;
    }
  }
}

void ComKInit(ll N){
  for(int i=0; i<=ComKTableLIMIT; i++){
    if(0<i && i<=N/2){
      ComK[i] = (ComK[i-1]*(N-i+1)/i)%MOD;
    }else if(i==0){
      ComK[i] = 1;
    }else if(N/2<i && i<=N){
      ComK[i] = ComK[N-i];
    }else{//i>Nの時
      ComK[i] = -1;
    }
  }
}

void ComInit(){//tableを作成
  FacMod[0] = FacMod[1] = 1;
  InvMod[0] = 0; InvMod[1] = 1; //InvMod[0]は定義できない
  FacInvMod[0] = FacInvMod[1] = 1;
  
  for(int i=2; i<LIMIT; i++){
    FacMod[i] = FacMod[i-1]*i%MOD;
    InvMod[i] = -InvMod[MOD%i]*(MOD/i)%MOD + MOD;
    FacInvMod[i] = FacInvMod[i-1] * InvMod[i]%MOD;
  }
}

ll ComMod(int n, int k){
  if(n>=0 && k>=0 && n>=k){
    return FacMod[n]*(FacInvMod[k]*FacInvMod[n-k]%MOD)%MOD;
  }else{
    return 0; //戻り値を0にする方が計算結果に影響が少ない。
  }
}


//繰り返し2乗法
ll power(ll x, ll y){//繰り返し2乗法
  if(y < 0){
    return 0;
  }else if(y == 0){
    return 1LL;
  }else if(y%2 == 0){//偶数
    return power(x,y/2)*power(x,y/2);
  }else{//奇数
    return power(x,y-1)*x;
  }
}

//繰り返し2乗法
ll mpower(ll x, ll y, ll m){//繰り返し2乗法
  if(y < 0){
    return 0;
  }else if(y == 0){
    return 1LL;
  }else if(y%2 == 0){//偶数
    return mpower(x,y/2,m)*mpower(x,y/2,m)%m;
  }else{//奇数
    return mpower(x,y-1,m)*x%m;
  }
}

//pair型のソート
bool compare_by_second(pair<ll,ll> a, pair<ll,ll> b){
  return a.second != b.second ? a.second < b.second : a.first < b.first;
}

class UnionFind {
  vector<int> parent;
  vector<int> rank;
  
  public:
  
  UnionFind(int N);
  int root(int x);
  void unite(int x, int y);
  bool isUnited(int x, int y);
  int getRank(int x);
};

UnionFind::UnionFind(int N):parent(N),rank(N,1){
  for(int i=0; i<N; i++){parent[i]=i;}
}
  
int UnionFind::root(int x){
  if(parent[x] == x) return x;
  return root(parent[x]);
  /*parent[x] = parent[parent[x]];
  if(parent[x]!=parent[parent[x]]) rank[x]-=1;*/
}
  
void UnionFind::unite(int x, int y) { // xとyの木を併合
  int rx = root(x); //xの根をrx
  int ry = root(y); //yの根をry
  if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
  
  if(rank[x] >= rank[y]){parent[ry] = rx; rank[y]+=1;}//大きい木に小さい木が併合される
  else{parent[rx] = ry; rank[x]+=1;}//根に直接辺を張る
}
  
bool UnionFind::isUnited(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
  int rx = root(x);
  int ry = root(y);
  parent[x]=rx; parent[y]=ry;rank[x]=2;rank[y]=2;
  return rx == ry;
}

int UnionFind::getRank(int x){
  return rank[x];
}

int main()
{
  //input
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  ll n, a, b; cin >> n >> a >> b;
  
  //main
  ll ans = mpower(2,n,MOD)-1;
  ComInit();
  ll na = 1;
  for(int i=1; i<=a; i++){
    na = (na*(n-i+1)%MOD*(InvMod[i])%MOD);
  }
  ll nb = 1;
  for(int i=1; i<=b; i++){
    nb = (nb*(n-i+1)%MOD*(InvMod[i])%MOD);
  }
  if(na < 0){
    na += MOD; 
  }
  if(nb < 0){
    nb += MOD; 
  }
  ans -= (na+nb);
  if(ans < 0){
    ans += MOD;
    if(ans < 0){
      ans += MOD; 
    }
  }
  
  // output
  cout << ans << endl;

  return 0;
}