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)

#include <bits/stdc++.h>
#define Int int64_t

using namespace std;

// [ _UnionFind ]
// Union-Find木 /* union by size */
// 宣言は UnionFind uf(N); のようにする
// find(x) : xの根を求める
// same(x, y) : xとyが同じ集合に属するかどうか
// size(x) : xの属する集合の要素数
// size() : 集合がいくつあるか
// unite(x, y) : xとyの属する集合を併合する
struct UnionFind {
private:
	vector<int> uf;
	size_t sz;

public:
	UnionFind(size_t Size) {
		uf.assign(Size, -1);
		sz = Size;
	}

	int find(int x) { return (uf[x] < 0) ? x : uf[x] = find(uf[x]); }
	bool same(int x, int y) { return find(x) == find(y); }
	int size(int x) { return -uf[find(x)]; }
	int size() { return sz; }

	void unite(int x, int y) {
		x = find(x), y = find(y);
		if (x == y) { return; }
		--sz;

		if (uf[y] < uf[x]) { swap(x, y); }
		uf[x] += uf[y];
		uf[y] = x;
	}
};

int main() {
	int N, M, K;
	cin >> N >> M >> K;
	UnionFind uf(N);
	vector<vector<int>> fr(N), bl(N);
	for (int i = 0; i < M; ++i) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		uf.unite(a, b);
		fr[a].emplace_back(b);
		fr[b].emplace_back(a);
	}
	for (int i = 0; i < K; ++i) {
		int a, b;
		cin >> a >> b;
		--a; --b;
		bl[a].emplace_back(b);
		bl[b].emplace_back(a);
	}

	vector<int> ans(N, 0);
	for (int i = 0; i < N; ++i) {
		ans[i] = uf.size(i) - fr[i].size() - 1;
		for (int v : bl[i]) {
			if (uf.same(i, v)) { --ans[i]; }
		}
	}
	for (int i = 0; i < N; ++i) {
		cout << ans[i] << (i < N - 1 ? " " : "\n");
	}

	return 0;
}