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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
using ll = long long;

struct UnionFind {
    UnionFind(int n) : p(n, -1) {}
    int root(int u) {
        return p[u] < 0 ? u : p[u] = root(p[u]);
    }
    int size(int u) {
        return -p[root(u)];
    }
    bool same(int u, int v) {
        return root(u) == root(v);
    }
    void unite(int u, int v) {
        u = root(u);
        v = root(v);
        if (u == v) return;
        if (p[u] > p[v]) swap(u, v);
        p[u] += p[v];
        p[v] = u;
    }
    vector<int> p;
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    int n, m, k;
    cin >> n >> m >> k;

    UnionFind uf(n);

    vector<int> r(n, 1);
    for (int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        a--; b--;

        uf.unite(a, b);
        r[a]++, r[b]++;
    }

    for (int i = 0; i < k; i++) {
        int c, d;
        cin >> c >> d;
        c--; d--;

        if (uf.same(c, d)) r[c]++, r[d]++;
    }

    for (int i = 0; i < n; i++) {
        cout << uf.size(i) - r[i] << " \n"[i == n - 1];
    }

    return 0;
}