描述：

C. Socks time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Arseniy is already grown-up and independent. His mother decided to leave him alone for m days and left on a vacation. She have prepared a lot of food, left some money and washed all Arseniy's clothes.

Ten minutes before her leave she realized that it would be also useful to prepare instruction of which particular clothes to wear on each of the days she will be absent. Arseniy's family is a bit weird so all the clothes is enumerated. For example, each of Arseniy's n socks is assigned a unique integer from 1

 to n. Thus, the only thing his mother had to do was to write down two integers li and ri for each of the days — the indices of socks to wear on the day i(obviously, li stands for the left foot and ri for the right). Each sock is painted in one of k colors.

When mother already left Arseniy noticed that according to instruction he would wear the socks of different colors on some days. Of course, that is a terrible mistake cause by a rush. Arseniy is a smart boy, and, by some magical coincidence, he posses k

 jars with the paint — one for each of k colors.

Arseniy wants to repaint some of the socks in such a way, that for each of m days he can follow the mother's instructions and wear the socks of the same color. As he is going to be very busy these days he will have no time to change the colors of any socks so he has to finalize the colors now.

The new computer game Bota-3 was just realised and Arseniy can't wait to play it. What is the minimum number of socks that need their color to be changed in order to make it possible to follow mother's instructions and wear the socks of the same color during each of m days.

Input

The first line of input contains three integers n, m and k (2 ≤ n ≤ 200 000, 0 ≤ m ≤ 200 000, 1 ≤ k ≤ 200 000) — the number of socks, the number of days and the number of available colors respectively.

The second line contain n integers c1, c2, ..., cn (1 ≤ ci ≤ k) — current colors of Arseniy's socks.

Each of the following m lines contains two integers li and ri (1 ≤ li, ri ≤ n, li ≠ ri) — indices of socks which Arseniy should wear during the i-th day.

Output

Print one integer — the minimum number of socks that should have their colors changed in order to be able to obey the instructions and not make people laugh from watching the socks of different colors.

Examples input

3 2 3
1 2 3
1 2
2 3

output

2

input

3 2 2
1 1 2
1 2
2 1

output

0

Note

In the first sample, Arseniy can repaint the first and the third socks to the second color.

In the second sample, there is no need to change any colors.

題意：

有n只襪子（1~n），k種顏色（1~k），在m天中，左腳穿下標為l，右腳穿下標為r的襪子，問最少修改幾隻襪子的顏色，可以使每天穿的襪子左右兩隻都同顏色。

思路一：

並查集處理出哪幾堆襪子是同一顏色的，對於每堆襪子求出出現最多顏色的次數，用這堆襪子的數目減去該值即為這堆襪子需要修改的顏色數，最後對每堆求和

程式碼一：

#include <bits/stdc++.h>
using  namespace  std;
#define rep(i,k,n) for(int i=k;i<=n;i++)

const int N=2e5+10;

int p[N], col[N], num[N];
int tol, ans;
std::vector<int> v[N];

int Find(int x){
  return p[x] == x ? x : p[x] = Find(p[x]);
}

void unite(int x, int y){
  x = Find(x);
  y = Find(y);
  if(x != y)
    p[y] = x;
}

int  main(){
  int n, m, k;
  cin >> n >> m >> k;
  rep(i, 1, n){
    cin >> col[i];
  } 
  rep(i, 1, n)p[i] = i; 
  int l, r; 
  rep(i, 1, m){
    cin >> l >> r;
    unite(l, r);//把顏色需要相同先放到一個集合中
  } 

  rep(i, 1, n){
    if(p[i] == i)num[i] = ++tol; // 給集合編號
  }
  rep(i, 1, n){
    v[num[Find(i)]].push_back(col[i]); //每個集合中加入顏色
  }
  rep(i, 1, tol){
    int sz = v[i].size();
    int mx = 0;
    std::map<int, int> mp;
    rep(j, 0, sz - 1){
      mp[v[i][j]] ++;
      mx = max(mx, mp[v[i][j]]);//找到集合中顏色最多的 把剩下的元素變成該顏色即可   
    }  
    ans += (sz - mx);
  }
  cout << ans << endl;
  return 0;
}

思路二：

每堆需要同色的襪子就是一個連通分支， 先建圖，在每個連通分支中，把所有襪子的顏色都改成出現顏色次數最多的那個顏色，即該分支節點個數 - 最多出現次數

程式碼二：

#include <bits/stdc++.h>
using  namespace  std;
#define rep(i,k,n) for(int i=k;i<=n;i++)

const int N=3e5+10;

std::vector<int> g[N];
int col[N], vis[N], cur, mx, ans;
std::map<int, int> num;

void dfs(int x){
  if(vis[x])return ;
  cur ++;
  mx = max(mx, ++num[col[x]]);
  vis[x] = 1;
  for(auto y : g[x]){
    dfs(y);
  }
}

int  main(){
  int n, m, k;
  cin >> n >> m >> k;
  rep(i, 1, n){
    cin >> col[i];
  }
  int l, r;
  rep(i, 1, m){
    cin >> l >> r;
    g[l].push_back(r);
    g[r].push_back(l);
  }
  rep(i, 1, n){
    if(! vis[i]){
      cur = 0;
      mx = 0;
      dfs(i);
      ans += (cur - mx);
      num.clear(); //不要忘了
    }
  }
  cout << ans << endl;
  return 0;
}