Codeforces 376C. Socks

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 l_i and r_i for each of the days — the indices of socks to wear on the day i (obviously, l_i stands for the left foot and r_i 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, ..., c_n (1 ≤ c_i ≤ k) — current colors of Arseniy's socks.

Each of the following m lines contains two integers l_i and r_i (1 ≤ l_i, r_i ≤ n, l_i ≠ r_i) — 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.

/*
这个人有n双袜子，每双袜子都有颜色编号，她妈妈出差但是给他规定了每天穿哪两只袜子，
如果这么穿的话，就可能穿的袜子不同颜色，他不想，所以他准备了k种颜料，准备把袜子颜色改了，问最少改多少次。

用并查集：将每天有联系的袜子（就是可能在同一天穿的袜子）构成树，然后组成森林，然后在一棵树上染袜子，将所有的袜子染成这个树上颜色
最多的那种颜色，然后累加
*/


/*
读错题了......呜呜呜自己眼瞎，最后还有十分钟刚有思路，就到时间了......呜呜呜，这才补出来
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<queue>
#include<map>
#include<cmath>
#define maxn 200005
#define LL long long
using namespace std;

int n,m,k,ans;
int a[maxn],f[maxn],w[maxn],v[maxn],vi[maxn];
vector<int> g[maxn];
int find(int x){return f[x]==x?x:f[x]=find(f[x]);}
int main(){
    scanf("%d%d%d",&n,&m,&k);
    for (int i=1;i<=n;i++) scanf("%d",&a[i]),f[i]=i;
    for (int i=1;i<=m;i++){int x,y;
        scanf("%d%d",&x,&y);
        if (find(x)!=find(y)){
            f[f[y]]=f[x];
        }
    }
    for (int i=1;i<=n;i++) g[find(i)].push_back(i);
    for (int i=1;i<=n;i++)
        if (f[i]==i){int p=g[i].size(),num=0,sum=0;
            for (int j=0;j<p;j++){
                if (!v[a[g[i][j]]]) v[a[g[i][j]]]=1,vi[++num]=a[g[i][j]];
                w[a[g[i][j]]]++;
            }
            for (int j=1;j<=num;j++) sum=max(sum,w[vi[j]]),v[vi[j]]=0,w[vi[j]]=0;
            ans+=p-sum;
        }
    printf("%d\n",ans);
    return 0;
}