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)

// Author: HarshKumar

#include <bits/stdc++.h>
using namespace std;
#define ll int64_t
const ll N = 1e6 + 6, M = 1e9 + 7;
struct DSU
{
    int parts;
    vector<int> rt, sz;

    DSU(int n)
    {
        rt.resize(n + 1);
        sz.resize(n + 1, 1);
        iota(rt.begin(), rt.end(), 0);
        parts = n;
    }

    int root(int k)
    {
        while (k != rt[k])
            k = rt[k] = rt[rt[k]];
        return k;
    }

    void U(int u, int v)
    {
        int r1 = root(u), r2 = root(v);
        if (r1 == r2)
            return;
        parts--;
        if (sz[r1] > sz[r2])
            swap(r1, r2);
        sz[r2] += sz[r1];
        sz[r1] = 0;
        rt[r1] = rt[r2];
    }
};
set<ll> g[N], h[N];
int main()
{
    ios::sync_with_stdio(false), cin.tie(nullptr);
    ll n, m, k, u, v;
    cin >> n >> m >> k;
    DSU d(n);
    while (m--)
        cin >> u >> v, u--, v--, g[u].insert(v), g[v].insert(u), d.U(u, v);
    while (k--)
        cin >> u >> v, u--, v--, h[u].insert(v), h[v].insert(u);
    for (ll i = 0; i < n; i++)
    {
        ll x = d.sz[d.root(i)] - 1 - g[i].size();
        for (ll e : h[i])
            if (d.root(e) == d.root(i) && g[i].find(e) == g[i].end())
                x--;
        cout << x << ' ';
    }
}