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++14 (GCC 5.4.1) (judge verdict: Accepted)

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

using lli = long long int;
using Vint = vector<int>;
using Vlli = vector<lli>;
using Wint = vector<Vint>;
using Wlli = vector<Vlli>;
using pii = pair<int, int>;
using pll = pair<lli, lli>;

const int MOD = 1e9 + 7;
const int INFi = 2e9 + 1;
const lli INFl = (lli)(9e18) + 1;
const vector<pii> DXDY = {make_pair(1, 0), make_pair(-1, 0), make_pair(0, 1), make_pair(0, -1)};

#define FOR(i, a, b) for(int (i) = (a); (i) < (b); (i)++)
#define FOReq(i, a, b) for(int (i) = (a); (i) <= (b); (i)++)
#define rFOR(i, a, b) for(int (i) = (b); (i) >= (a); i--)
#define FORstep(i, a, b, step) for(int (i) = (a); i < (b); i += (step))
#define REP(i, n) FOR(i, 0, n)
#define rREP(i, n) rFOR(i, 0, (n-1))
#define vREP(i, vec) for(auto &(i) : (vec))


template <class T> inline int argmin(vector<T> vec){return min_element(vec.begin(), vec.end()) - vec.begin();}
template <class T> inline int argmax(vector<T> vec){return max_element(vec.begin(), vec.end()) - vec.begin();}
template <class T> inline void chmax(T &a, T b){a = max(a, b);}
template <class T> inline void chmin(T &a, T b){a = min(a, b);}
inline int toInt(string &s){int res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
inline int toInt(const string s){int res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
inline long long int toLong(string &s){lli res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
inline long long int toLong(const string s){lli res = 0; for(char a : s) res = 10 * res + (a - '0'); return res;}
template <class T> inline std::string toString(T n){
  if(n == 0) return "0";
  std::string res = "";
  if(n < 0){n = -n;while(n != 0){res += (char)(n % 10 + '0'); n /= 10;} std::reverse(res.begin(), res.end()); return '-' + res;}
  while(n != 0){res += (char)(n % 10 + '0'); n /= 10;} std::reverse(res.begin(), res.end()); return res;
}

// ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


int main(void){
  lli res = 0;
  int a, n; cin >> n;
  map<int, int> data;
  REP(i, n){
    cin >> a;
    data[a]++;
  }
  vREP(pr, data){
    if(pr.first > pr.second) res += pr.second;
    else res += (pr.second - pr.first);
  }
  cout << res << '\n';
  return 0;
}