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;
#define fi first
#define se second
#define rep(i,n) for(int i=0, i##_len=(n); i<i##_len; ++i)
#define rrep(i,n) for(int i=1,i##_len=(n); i<=i##_len; ++i)
#define drep(i,n) for(int i=(n)-1; i>=0; --i)
#define srep(i,s,t) for(int i=(s);i<(t);++i)
#define rng(obj) (obj).begin(), (obj).end()
#define maxs(x,y) (x=max((x),(y)))
#define mins(x,y) (x=min((x),(y)))
#define limit(x,l,r) max(l,min(x,r))
#define lims(x,l,r) (x=max(l,min(x,r)))
#define isin(x,l,r) ((l) <= (x) && (x) < (r))
#define pb push_back
#define sz(x) (int)(x).size()
#define uni(x) x.erase(unique(rng(x)),x.end())
#define show(x) cout<<#x<<" = "<<x<<endl;
#define PQ(T) priority_queue<T,v(T),greater<T>>
#define bn(x) ((1<<x)-1)
#define dup(x,y) (((x)+(y)-1)/(y))
#define newline puts("")
#define v(T) vector<T>
#define vv(T) vector<vector<T>>
using ll = long long;
using ull = unsigned long long;
using uint = unsigned int;
using P = pair<int,int>;
using vi = vector<int>;
using vvi = vector<vi>;
using vl = vector<ll>;
using vp = vector<P>;
template <class T> inline istream& operator>>(istream&i,v(T)&v)
{ rep(j,sz(v))i>>v[j];return i;}
template <class T> string join(const v(T)&v)
{ stringstream s;rep(i,sz(v))s<<' '<<v[i];return s.str().substr(1);}
template <class T> inline ostream& operator<<(ostream&o,const v(T)&v)
{ if(sz(v))o<<join(v); return o;}
template <class T1,class T2> inline istream& operator>>(istream&i ,pair<T1,T2>&v)
{ return i>>v.fi>>v.se;}
template <class T1,class T2> inline ostream& operator<<(ostream&o ,pair<T1,T2>&v)
{ return o<<v.fi<<","<<v.se;}
template <class T> inline ll suma(const v(T)& a) 
{ ll res(0); for(auto&& x:a) res += x; return res;}
const int INF = 1001001001;
const ll LINF = 1001002003004005006ll;
const double eps = 1e-10;
const int MX = 200005;
//----------------------------------------------------------------

#define fileio ifstream fio; if(argc>=2){fio.open(argv[1]); std::cin.rdbuf(fio.rdbuf());}
#define fastio ios_base::sync_with_stdio(false);
#define MOD 1000000007
//----------------------------------------------------------------

void Main(){
  int n;
  cin>>n;
  vl a(n);
  cin>>a;
  unordered_map<ll,ll> m;
  rep(i,n){
    m[a[i]]++;
  }
  ll ans = 0;
  for(auto i:m){
    if(i.se>i.fi) ans+=i.se-i.fi;
    else if(i.se<i.fi) ans+=i.se;
  }
  cout<<ans<<endl;
} int main(int argc,char* argv[]){ 
    fastio; fileio; Main();}