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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

/*
このコード、と～おれ!
Be accepted!
∧＿∧　
（｡･ω･｡)つ━☆・*。
⊂　　 ノ 　　　・゜+.
　しーＪ　　　°。+ *´¨)
 　　　　　　　　　.· ´¸.·*´¨) ¸.·*¨)
		  　　　　　　　　　　(¸.·´ (¸.·'* ☆
*/

#include <cstdio>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstring>
#include <vector>
#include <numeric>
#include <iostream>
#include <random>
#include <map>
#include <unordered_map>
#include <queue>
#include <regex>
#include <functional>
#include <complex>
#include <list>
#include <cassert>
#include <iomanip>
#include <set>
/*多倍長整数/cpp_intで宣言
#include <boost/multiprecision/cpp_int.hpp>
using namespace boost::multiprecision;
*/

//#pragma gcc target ("avx2")
//#pragma gcc optimization ("o3")
//#pragma gcc optimization ("unroll-loops")
#define rep(i, n) for(int i = 0; i < (n); ++i)
#define rep1(i, n) for(int i = 1; i <= (n); ++i)
#define rep2(i, n) for(int i = 2; i < (n); ++i)
#define repr(i, n) for(int i = n; i >= 0; --i)
#define reprm(i, n) for(int i = n - 1; i >= 0; --i)
#define printynl(a) printf(a ? "yes\n" : "no\n")
#define printyn(a) printf(a ? "Yes\n" : "No\n")
#define printYN(a) printf(a ? "YES\n" : "NO\n")
#define printim(a) printf(a ? "possible\n" : "imposible\n")
#define printdb(a) printf("%.50lf\n", a) //少数出力
#define printdbd(a) printf("%.16lf\n", a) //少数出力(桁少なめ)
#define prints(s) printf("%s\n", s.c_str()) //string出力
#define all(x) (x).begin(), (x).end()
#define allsum(a, b, c) ((a + b) * c / 2.0) //等差数列の和、初項,末項,項数
#define pb push_back
#define priq priority_queue
#define rpriq priq<int, vector<int>, greater<int>>
#define deg_to_rad(deg) (((deg)/360.0)*2.0*PI)
#define rad_to_deg(rad) (((rad)/2.0/PI)*360.0)
#define Please return
#define AC 0
#define addf(T) [](T a, T b){return (a + b);}
#define minf(T) [](T a, T b){return min(a, b);}
#define maxf(T) [](T a, T b){return max(a, b);}
#define nopf(T) [](T a){return a;}
#define manhattan_dist(a, b, c, d) (abs(a - c) + abs(b - d)) /*(a, b) から (c, d) のマンハッタン距離 */


using ll = long long;

constexpr int INF = 1073741823;
constexpr int MINF = -1073741823;
constexpr ll LINF = ll(4661686018427387903);
constexpr ll MOD = 1000000007;
const long double PI = acos(-1.0L);

using namespace std;

void scans(string& str) {
	char c;
	str = "";
	scanf("%c", &c);
	if (c == '\n')scanf("%c", &c);
	while (c != '\n' && c != -1 && c != ' ') {
		str += c;
		scanf("%c", &c);
	}
}

void scanc(char& str) {
	char c;
	scanf("%c", &c);
	if (c == -1)return;
	while (c == '\n') {
		scanf("%c", &c);
	}
	str = c;
}

double acot(double x) {
	return PI / 2 - atan(x);
}

ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a % b);
}

ll lcm(ll number1, ll number2) {
	return number1 / gcd(number1, number2) * number2;
}

ll LSB(ll n) { return (n & (-n)); }

/*-----------------------------------------ここからコード-----------------------------------------*/

//0-indexed
struct unionfind {
	vector<ll> par, siz;

	unionfind(ll n) : par(n), siz(n) {
		for (ll i = 0; i < n; ++i) {
			//全部根で初期化
			par[i] = i;
			//サイズは1
			siz[i] = 1;
		}
	}

	void init(ll n) {
		par.resize(n);
		siz.resize(n);
		for (ll i = 0; i < n; ++i) {
			//全部根で初期化
			par[i] = i;
			//サイズは1
			siz[i] = 1;
		}
	}

	//根を返す
	ll find(ll a) {
		return par[a] == a ? a : par[a] = find(par[a]);
	}

	//くっつける。元から同じだったらfalseを返す
	bool unite(ll a, ll b) {
		ll x = find(a), y = find(b);
		if (x == y)return false;
		else if (siz[x] < siz[y]) {
			par[x] = y;
			siz[y] += siz[x];
		}
		else if (siz[x] > siz[y]) {
			par[y] = x;
			siz[x] += siz[y];
		}
		else {
			par[x] = y;
			siz[y] += siz[x];
		}
		return true;
	}

	//同じ集合か判定する
	bool same(ll a, ll b) {
		return find(a) == find(b);
	}

	//サイズを返す
	ll size(ll a) {
		return siz[find(a)];
	}

	//同じ集合に属す葉を纏めて返す
	vector<ll> leaf(ll a) {
		vector<ll> x;
		ll n = par.size();
		for (ll i = 0; i < n; ++i)if (same(a, i))x.pb(i);
		return x;
	}

};

int main() {

	int n, m, k;
	scanf("%d%d%d", &n, &m, &k);
	unionfind fri(n);
	vector<vector<bool>> fr(n, vector<bool>(n)), bl(n, vector<bool>(n));
	int a, b;
	rep(i, m) {
		scanf("%d%d", &a, &b);
		--a; --b;
		fr[a][b] = fr[b][a] = true;
		fri.unite(a, b);
	}
	rep(i, k) {
		scanf("%d%d", &a, &b);
		--a; --b;
		bl[a][b] = bl[b][a] = true;
	}
	vector<int> ans(n);
	rep(i, n) {
		for(int j = i + 1; j < n; ++j) {
			if (!fr[i][j] && fri.same(i, j) && !bl[i][j]) {
				++ans[i];
				++ans[j];
			}
		}
	}
	for (const auto& aa : ans)printf("%d\n", aa);

	Please AC;
}