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>
using namespace std;
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return 1;
    }
    return 0;
}
typedef long long int ll;

#define ALL(v) (v).begin(), (v).end()
#define RALL(v) (v).rbegin(), (v).rend()
#define endl "\n"
const double EPS = 1e-7;
const int INF = 1 << 30;
const ll LLINF = 1LL << 60;
const double PI = acos(-1);
const int MOD = 1000000007;
const int dx[4] = {1, 0, -1, 0};
const int dy[4] = {0, 1, 0, -1};

//-------------------------------------

using P = pair<int, int>;

struct UnionFind {
    vector<int> par;

    UnionFind(int n) : par(n, -1) {}
    void init(int n) { par.assign(n, -1); }

    int root(int x) {
        if(par[x] < 0)
            return x;
        else
            return par[x] = root(par[x]);
    }

    bool issame(int x, int y) { return root(x) == root(y); }

    bool merge(int x, int y) {
        x = root(x);
        y = root(y);
        if(x == y)
            return false;
        if(par[x] > par[y])
            swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    int size(int x) { return -par[root(x)]; }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, m, k;
    cin >> n >> m >> k;
    UnionFind uf(n);
    vector<int> ans(n, 0);
    vector<vector<int>> g(n);
    for(int i = 0; i < m; i++) {
        int a, b;
        cin >> a >> b;
        a--;
        b--;
        g[a].push_back(b);
        g[b].push_back(a);
        uf.merge(a, b);
    }
    for(int i = 0; i < n; i++) {
        ans[i] += uf.size(i);
        ans[i] -= ((int)g[i].size() + 1);
    }
    for(int i = 0; i < k; i++) {
        int c, d;
        cin >> c >> d;
        c--;
        d--;
        if(uf.issame(c, d)) {
            ans[c]--;
            ans[d]--;
        }
    }
    for(int i = 0; i < n; i++) {
        cout << ans[i] << ' ';
    }
    cout << endl;
}