Problem:
Score : 400 points

Problem Statement

An SNS has N users - User 1, User 2, \cdots, User N.

Between these N users, there are some relationships - M friendships and K blockships.

For each i = 1, 2, \cdots, M, there is a bidirectional friendship between User A_i and User B_i.

For each i = 1, 2, \cdots, K, there is a bidirectional blockship between User C_i and User D_i.

We define User a to be a friend candidate for User b when all of the following four conditions are satisfied:

a \neq b.

There is not a friendship between User a and User b.

There is not a blockship between User a and User b.

There exists a sequence c_0, c_1, c_2, \cdots, c_L consisting of integers between 1 and N (inclusive) such that c_0 = a, c_L = b, and there is a friendship between User c_i and c_{i+1} for each i = 0, 1, \cdots, L - 1.

For each user i = 1, 2, ... N, how many friend candidates does it have?

Constraints

All values in input are integers.

2 ≤ N ≤ 10^5

0 \leq M \leq 10^5

0 \leq K \leq 10^5

1 \leq A_i, B_i \leq N

A_i \neq B_i

1 \leq C_i, D_i \leq N

C_i \neq D_i

(A_i, B_i) \neq (A_j, B_j) (i \neq j)

(A_i, B_i) \neq (B_j, A_j)

(C_i, D_i) \neq (C_j, D_j) (i \neq j)

(C_i, D_i) \neq (D_j, C_j)

(A_i, B_i) \neq (C_j, D_j)

(A_i, B_i) \neq (D_j, C_j)

Input

Input is given from Standard Input in the following format:

N M K
A_1 B_1
\vdots
A_M B_M
C_1 D_1
\vdots
C_K D_K

Output

Print the answers in order, with space in between.

Sample Input 1

4 4 1
2 1
1 3
3 2
3 4
4 1

Sample Output 1

0 1 0 1

There is a friendship between User 2 and 3, and between 3 and 4. Also, there is no friendship or blockship between User 2 and 4. Thus, User 4 is a friend candidate for User 2.

However, neither User 1 or 3 is a friend candidate for User 2, so User 2 has one friend candidate.

Sample Input 2

5 10 0
1 2
1 3
1 4
1 5
3 2
2 4
2 5
4 3
5 3
4 5

Sample Output 2

0 0 0 0 0

Everyone is a friend of everyone else and has no friend candidate.

Sample Input 3

10 9 3
10 1
6 7
8 2
2 5
8 4
7 3
10 9
6 4
5 8
2 6
7 5
3 1

Sample Output 3

1 3 5 4 3 3 3 3 1 0

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

/*
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@TT_beginner
*/
#include "bits/stdc++.h"
using namespace std;
#define ok1 printf("ok1\n")
#define ok2 printf("ok2\n")
#define M LLONG_MAX
#define rep(i,n) for(int i=0;i<n;++i)
#define REP(i,s,n) for(int i=(s);i<(n);++i)
#define repr(i,n) for(int i=n-1;i>=0;--i)
#define REPR(i,s,n) for(int i=(s);i>=(n);--(i))
#define all(a) (a).begin(),(a).end()
#define reall(a) (a).rbegin(),(a).rend()
#define pb emplace_back //emplace_backの方が速いが使い慣れてないため
#define DOUBLE fixed << setprecision(15)
#define fi first
#define se second
#define mp make_pair
#define mt make_tuple
const double pi = acos(-1.0L);
typedef vector<int> vi;
typedef vector<string> vs;
typedef long long ll;
typedef vector<ll> vll;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
typedef vector<char> vc;
typedef vector<double> vd;
typedef vector<bool> vb;
typedef deque<ll> dll;
typedef pair<ll, ll> P;
typedef tuple<ll, ll, ll> TLL;
typedef vector<P> vP;
typedef vector<TLL> vTLL;
const ll mod = 1e9 + 7;
//const ll mod = 998244353;
ll dy[4] = { 1,0,-1,0 };
ll dx[4] = { 0,1,0,-1 };
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }
template<class A, class B>
ostream& operator<<(ostream& ost, const pair<A, B>&p) {
	ost << "{" << p.first << ", " << p.second << "} ";
	return ost;
}
template<class T>
ostream& operator<<(ostream& ost, const vector<T>&v) {
	ost << "{";
	for (int i = 0; i<v.size(); i++) {
		if (i)ost << ", ";
		ost << v[i];
	}
	ost << "} ";
	return ost;
}
template<class A, class B>
ostream& operator<<(ostream& ost, const map<A, B>&v) {
	ost << "{";
	for (auto p : v) {
		ost << "{" << p.first << ", " << p.second << "} ";
	}
	ost << "} ";
	return ost;
}
bool out_check(ll a, ll b) { return (0 <= a && a < b); }
double pitagoras(ll a, ll b, ll c, ll d) {
	double dx = a - b, dy = c - d;
	return pow(dx * dx + dy * dy, 0.5);
}
ll modpow(ll a, ll b) { ll c = 1; while (b > 0) { if (b & 1) { c = a * c%mod; }a = a * a%mod; b >>= 1; }return c; }
void Yes(bool x) { cout << ((x) ? "Yes\n" : "No\n"); }
void YES(bool x) { cout << ((x) ? "YES\n" : "NO\n"); }
void yes(bool x) { cout << ((x) ? "yes\n" : "no\n"); }
void Yay(bool x) { cout << ((x) ? "Yay!\n" : ":(\n"); }
ll  n, m, d, e, r, l, k, h, w, Q, ans, ret = M;
bool flag = false, flag2 = false, flag3 = false;
string s, t, u;
ll x, y;
vvll g, block;
struct UnionFind {
	vector<long long> data;
	UnionFind(long long size) : data(size, -1) { }
	bool unite(long long x, long long y) {
		x = root(x); y = root(y);
		if (x != y) {
			if (data[y] < data[x]) swap(x, y);
			data[x] += data[y]; data[y] = x;
		}
		return x != y;
	}
	bool same(long long x, long long y) {
		return root(x) == root(y);
	}
	int root(long long x) {
		return data[x] < 0 ? x : data[x] = root(data[x]);
	}
	int size(long long x) {
		return -data[root(x)];
	}
};

int main() {
	cin.tie(0); ios::sync_with_stdio(false);//not intractive
	cin >> n >> m >> k;
	g.resize(n), block.resize(n);
	UnionFind tree(n);
	rep(i, m) {
		cin >> x >> y;
		x--, y--;
		g[x].pb(y);
		g[y].pb(x);
		tree.unite(x, y);
	}
	rep(i, k) {
		cin >> x >> y;
		x--, y--;
		block[x].pb(y);
		block[y].pb(x);
	}
	rep(i, n) {
		ans = tree.size(i)-g[i].size()-1;
		for (auto p : block[i]) if (tree.same(i, p)) ans--;
		cout << ans << " ";
	}
	return 0;
}