Problem:
Score : 300 points

Problem Statement

You are given a sequence of positive integers of length N, a = (a_1, a_2, ..., a_N).
Your objective is to remove some of the elements in a so that a will be a good sequence.

Here, an sequence b is a good sequence when the following condition holds true:

For each element x in b, the value x occurs exactly x times in b.

For example, (3, 3, 3), (4, 2, 4, 1, 4, 2, 4) and () (an empty sequence) are good sequences, while (3, 3, 3, 3) and (2, 4, 1, 4, 2) are not.

Find the minimum number of elements that needs to be removed so that a will be a good sequence.

Constraints

1 \leq N \leq 10^5

a_i is an integer.

1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 ... a_N

Output

Print the minimum number of elements that needs to be removed so that a will be a good sequence.

Sample Input 1

4
3 3 3 3

Sample Output 1

1

We can, for example, remove one occurrence of 3. Then, (3, 3, 3) is a good sequence.

Sample Input 2

5
2 4 1 4 2

Sample Output 2

2

We can, for example, remove two occurrences of 4. Then, (2, 1, 2) is a good sequence.

Sample Input 3

6
1 2 2 3 3 3

Sample Output 3

0

Sample Input 4

1
1000000000

Sample Output 4

1

Remove one occurrence of 10^9. Then, () is a good sequence.

Sample Input 5

8
2 7 1 8 2 8 1 8

Sample Output 5

5

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

const int64_t MOD = 1000000007;
typedef int64_t ll;
typedef uint64_t ull;

#define FOR(i, start, end) for(uint64_t i=start; i<end; i++)
#define REP(i, n) FOR(i, 0, n)

// 最大公約数gcd
// 最小公倍数lcm=m*n/gcd
uint64_t gcd(uint64_t m, uint64_t n) {
  uint64_t temp;
  while (m % n != 0){
    temp = n;
    n = m % n;
    m = temp;
  }
  return n;
}

uint64_t lcm(uint64_t m, uint64_t n) {
  return (m*n)/gcd(m,n);
}

// vector<vector<uint64_t> > v(n+1, vector<uint64_t>(n+1, 0))
// v[n][k]に組み合わせ数が入る。
void comb(vector<vector <uint64_t> > &v){
  for(uint64_t i = 0;i <v.size(); i++){
    v[i][0]=1;
    v[i][i]=1;
  }
  for(uint64_t k = 1;k <v.size();k++){
    for(uint64_t j = 1;j<k;j++){
      v[k][j]=(v[k-1][j-1]+v[k-1][j]);
    }
  }
}

// 掛け算オーバーフロー判定
bool is_product_overflow(uint64_t a, uint64_t b) {

	uint64_t prod = a * b;
	
	return (prod / b != a);
}

//素因数分解
void primeFactorization(uint64_t a, list<uint64_t> &factors){ //素因数分解を出力するプログラム
  long i,sq;
  if(a%2==0){ //偶数の場合
    factors.push_back(2);
    primeFactorization(a/2,factors); //2で割った値で再帰
    return;
  }
  sq = sqrt(a);
  for(i=3;i<=sq;i+=2){ //3以上√a以下の奇数の場合
	if(a%i==0){
	  factors.push_back(i);
      primeFactorization(a/i,factors); //割れた値で再帰
      return;
	}
  }
  //偶数でも3以上√a以下の奇数の場合でも割り切れない場合
  if(a!=1){ //aが1でないなら、a自身は素数
    factors.push_back(a);
  }
}

// フェルマーの小定理
// mod. m での a の逆元 a^{-1} を計算する
// (a/b)%m = a*movinv(b,m)%m  
int64_t modinv(int64_t a, int64_t m) {
    int64_t b = m, u = 1, v = 0;
    while (b) {
        int64_t t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

// 円周率
// M_PI

// #include <iomanip> // setprecisionを使用するのに必要
// cout << std::fixed << std::setprecision(15) << y << endl;

// 昇順
// priority_queue<int, vector<int>, greater<int> > queue;

// 降順ソート vector<int> v;
// sort(v.begin(),v.end(),greater<int>());

signed main() {
  
  int n;
  cin >> n;
  
  unordered_map<int, int> count;
  
  int a;
  for(int i=0;i<n;i++){
    cin >> a;
    count[a]++;
  }
  
  int countToBeRemoved = 0;
  for(auto p:count){
    if(p.first<p.second){
      countToBeRemoved += p.second-p.first;
    }
    else if(p.first>p.second){
      countToBeRemoved += p.second;
    }
  }
  cout << countToBeRemoved << endl; 
  return 0;
}