Initially, each sock is its own set.
If on a certain day the two socks to be worn are $l_i$ and $r_i$, then merge the sets containing $l_i$ and $r_i$
It’s evident that ultimately, all socks in each set must be of the same color.
Using union find set, find the size of each set $|S|$, as well as the count of the most frequently occurring sock color in the set $s_{\max}$. The rest need to be repainted. Sum up the answers. If using a map, the complexity is $O(n\log)$.

The second solution is constructing an undirected graph, using DFS on each connected component and maintaining it with a hash table, with an expected complexity of $O(n)$.

// compile: make data
// run: ./data < data.in
#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#ifdef LOCAL
#include <debug/codeforces.h>
#define debug(x...) _debug_print(#x, x);
#define Debug(x...) _debug_print_format(#x, x);
std::ifstream terminal("/dev/tty");
#define PP cerr<<"\033[1;30mpause...\e[0m",terminal.ignore();
#else
#define debug(x...)
#define Debug(x...)
#define PP
#endif
template<typename...Args> void print_(Args...args){((cout<<args<<" "),...)<<endl;}
#define rep(i,a,b) for(int i=(a);i<(int)(b);++i)
#define sz(v) ((int)(v).size())
#define print(...) print_(__VA_ARGS__);
#define FIND(a, x) ((find(a.begin(),a.end(),(x))!=a.end())?1:0)
#define cmin(x,...) x=min({(x),__VA_ARGS__})
#define cmax(x,...) x=max({(x),__VA_ARGS__})
#define INTMAX (int)(9223372036854775807)
#define INF (int)(1152921504606846976)
#define NaN (int)(0x8b88e1d0595d51d1)
#define double long double
#define int long long
#define uint unsigned long long
#define MAXN 200010

struct DSU {
    vector<int> parent, rank, s;
    DSU(int n) {
        parent.resize(n);
        rank.resize(n);
        s.resize(n, 1);
        rep(i, 0, n) parent[i] = i;
    }
    int find(int x) {
        if (parent[x] != x) parent[x] = find(parent[x]);
        return parent[x];
    }
    void unite(int x, int y) {
        int rx = find(x), ry = find(y);
        if (rx == ry) return;
        if (rank[rx] < rank[ry]) parent[rx] = ry, s[ry] = s[rx] + s[ry];
        else if (rank[rx] > rank[ry]) parent[ry] = rx, s[rx] = s[rx] + s[ry];
        else parent[ry] = rx, ++rank[rx], s[rx] = s[rx] + s[ry];
    }
    int get_size(int x) {
        return s[find(x)];
    }
};

int32_t main() {
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);

    int n, m, k; cin >> n >> m >> k;
    DSU dsu(n+1);
    vector<int> a(n);
    vector<vector<int>> v(n);
    rep(i, 0, n) cin >> a[i];
    rep(i, 0, m) {
        int x, y; cin >> x >> y;
        --x, --y;
        dsu.unite(x, y);
    }
    rep(i, 0, n) v[dsu.find(i)].push_back(a[i]);
    int ans = 0;
    rep(i, 0, n) {
        if (sz(v[i]) > 1) {
            map<int, int> mp;
            int maxv = 0;
            for (auto j: v[i]) {
                mp[j]++;
                cmax(maxv, mp[j]);
            }
            ans += sz(v[i]) - maxv;
        }
    }
    cout << ans << endl;

    return 0;
}