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// Author : Jakub Rożek
// Task   : Liderzy
// Memory : O(n)
// Time   : O(n)
// Solv   : Rozwiązanie wzorcowe

#include "bits/stdc++.h"
using namespace std;
using LL = long long;
template <typename T>
using P = pair<T, T>;
template <typename T>
using VV = vector<vector<T>>;
#define all(x) x.begin(), x.end()
#define FOR(i,a,b) for(int i=(a); i<=(b); ++i)
#define FORD(i,a,b) for(int i=(a); i>=(b); --i)
#define REP(i,n) for(int i=0; i<(n); ++i)
#define ssize(x) int((x).size())
#ifdef DEBUG
template <typename T1, typename T2>
auto&operator<<(auto&o,pair<T1,T2>p){return o<<'('<<p.first<<", "<<p.second<<")";}
auto operator<<(auto&o,auto x)->decltype(x.end(),o){o<<"{";for(auto e:x)o<<","<<e;return o<<"}";}
#define debug(x...) cerr<<"["#x"]: ",[](auto...$){((cerr<<$<<"; "),...)<<endl;}(x)
#else
#define debug(...) {}
#endif

const int N = 500'000;

int n, x, moge, mam, odp;
int ile[N+1]; // ile mamy jakich liczb.
int tab[N+1]; // ile mamy grup danej wielkosci.

void solution() {
    cin >> n;
    FOR (i, 1, n) {
        cin >> x;
        ile[x]++;
    }
    FOR (i, 1, n) {
        tab[ile[i]]++;
    }
    
    odp = 0;
    mam = n;
    moge = 0;
    FORD (i, n, 0) {
        if (tab[i] == 0) continue;
        REP (j, tab[i]) {
            ++odp;
            mam -= i;
            moge += i-1;
            if (moge >= mam) {
                cout << odp << '\n';
                return;
            }
        }
    }
}

int main() {
    cin.tie(0)->sync_with_stdio(0);
    int tests = 1;
    // cin>>tests;
    FOR (i,1,tests) {
        solution();
    }    
    return 0;
}