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: Time Limit Exceeded)

#include <bits/stdc++.h>
#define rep(i, n) for (int i = 0; i < (n); ++i)
using namespace std;
//using ll = long long;
using P = pair<int, int>;

int main() {
    int n, m, k;
    cin >> n >> m >> k;
    vector<P> p(m);
    vector<P> fbNum(n);     // first:friend, second:block
    rep(i,m) {
        int a, b;
        cin >> a >> b;
        if (a > b) swap(a,b);
        a--; b--;
        p[i].first = a;
        p[i].second = b;
        fbNum[a].first++; fbNum[b].first++;
    }
    // グループ分け
    vector<int> group(n);
    vector<int> gNum;
    int id = 0;
    rep(i,n) {
        if (group[i] != 0) continue;
        id++;
        gNum.push_back(0);
        queue<int> q;
        q.push(i);
        group[i] = id;
        while (!q.empty()) {
            int node = q.front(); q.pop();
            gNum[id-1]++;
            rep(j,m) {
                if (p[j].first == node && group[p[j].second] == 0) {
                    q.push(p[j].second);
                    group[p[j].second] = id;
                }
                else if (p[j].second == node && group[p[j].first] == 0) {
                    q.push(p[j].first);
                    group[p[j].first] = id;
                }
            }
        }
    }
    rep(i,k) {
        int a, b;
        cin >> a >> b;
        a--; b--;
        if (group[a] == group[b]){
            fbNum[a].second++;
            fbNum[b].second++;
        }
    }
    
    rep(i,n) {
        int ans = gNum[group[i]-1] - fbNum[i].first - fbNum[i].second - 1;
        cout << ans << " ";
    }
    return 0;
}