human
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 ≠...
Solve the task in PYTHON3.
assistant
#Here 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.

import math
import time
from collections import defaultdict,deque,Counter
from sys import stdin,stdout
from bisect import bisect_left,bisect_right
from queue import PriorityQueue 
import sys
class Graph:
    def __init__(self, n):
        self.graph = defaultdict(lambda: [])
        self.vertices = n
        self.edges = 0
        self.toposorted = []

    def addEdge(self, a, b):  # tested
        self.graph[a].append(b)
        self.edges+=1

    def cycleUntill(self, visited, curr, parent):  # tested
        visited[curr] = True
        for i in self.graph[curr]:
            if(not visited[i]):
                if(self.cycleUntill(visited, i, curr)):
                    return True
            elif(i != parent):
                return True
        return False

    def cycle(self):  # tested
        n = self.vertices
        visited = [False]*(n+2)
        for i in range(1, n+1):
            if(not visited[i]):
                if(self.cycleUntill(visited, i, -1)):
                    return True
        return False

    def topologicalSort(self):#tested
        in_degree = [0]*(self.vertices+10)
        for i in self.graph:
            for j in self.graph[i]:
                in_degree[j] += 1
        queue = deque()
        for i in range(1,self.vertices+1):
            if in_degree[i] == 0:
                queue.append(i)
        cnt = 0
        while queue:
            u = queue.popleft()
            self.toposorted.append(u)
            for i in self.graph[u]:
                in_degree[i] -= 1
                if in_degree[i] == 0:
                    queue.append(i)
            cnt += 1
        if cnt != self.vertices:
            return False
        else:
            return True
    def connected(self):
        visited=[False]*(self.vertices +2)
        ans=[]
        for i in range(1,self.vertices+1):
            if(not visited[i]):
                comp=[]
                q=deque()
                visited[i]=True
                q.append(i)
                while(len(q)>0):
                   ...