from itertools import permutations
import sys

sys.setrecursionlimit(10 ** 6)
from bisect import *
from collections import *
from heapq import *

def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def SI(): return sys.stdin.readline()[:-1]
def LLI(rows_number): return [LI() for _ in range(rows_number)]
def LLI1(rows_number): return [LI1() for _ in range(rows_number)]
int1 = lambda x: int(x) - 1
def MI1(): return map(int1, sys.stdin.readline().split())
def LI1(): return list(map(int1, sys.stdin.readline().split()))
p2D = lambda x: print(*x, sep="\n")
dij = [(1, 0), (0, 1), (-1, 0), (0, -1)]

class UnionFind:
    def __init__(self, n):
        self.state = [-1] * n
        self.member=[[i] for i in range(n)]

    def root(self, u):
        v = self.state[u]
        if v < 0: return u
        self.state[u] = res = self.root(v)
        return res

    def merge(self, u, v):
        res=[]
        ru = self.root(u)
        rv = self.root(v)
        if ru == rv: return res
        if ru > rv: ru, rv = rv, ru
        if ru==0:res=self.member[rv]
        else:self.member[ru]+=self.member[rv]
        self.state[rv] = ru
        return res

def main():
    # 橋を壊していくのではなく、クエリを逆順にして、橋をつないでいく
    n,m,q=MI()
    ab=LLI1(m)
    cd=LLI1(q)
    uf=UnionFind(n)
    # 破壊される橋をチェック
    des=[[False]*n for _ in range(n)]
    for c,d in cd:des[c][d]=des[d][c]=True
    # 破壊されない橋を結ぶ
    ans=[0]*n
    for a,b in ab:
        if des[a][b]:continue
        uu=uf.merge(a,b)
        for u in uu:ans[u]=-1
    # クエリを逆から見て橋をつないでいく
    for i,(c,d) in enumerate(cd[::-1]):
        uu=uf.merge(c,d)
        for u in uu:ans[u]=q-i
    for i in range(1,n):print(ans[i])

main()