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;

#define MEM(a, b) memset(a, (b), sizeof(a))
#define repn(i, a, n) for (int i = a; i < n; i++)
#define revn(i, a, n) for (int i = a; i > n; i--)
#define rep(i, n) for (int i = 0; i < n; i++)
#define rev(i, a) for (int i = a; i > -1; i--)
#define mp make_pair
#define pb push_back
#define INF 1000000010
#define LLINF 1000000000000000010;
#define MOD 1000000007 // 998244353 1000000007
#define PI 3.141592653589793238462643383279502884197169399375105820974945
#define fastio ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
#define all(v) v.begin(), v.end()
typedef long long ll;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<vi> vii;
typedef pair<int, int> pii;

//debugging output in cerr
#define n_l '\n'
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]: "; cerr << to_string(__VA_ARGS__) << endl
template <typename T, size_t N> int SIZE(const T (&t)[N]){ return N; } template<typename T> int SIZE(const T &t){ return t.size(); } string to_string(const string s, int x1=0, int x2=1e9){ return '"' + ((x1 < s.size()) ? s.substr(x1, x2-x1+1) : "") + '"'; } string to_string(const char* s) { return to_string((string) s); } string to_string(const bool b) { return (b ? "true" : "false"); } string to_string(const char c){ return string({c}); } template<size_t N> string to_string(const bitset<N> &b, int x1=0, int x2=1e9){ string t = ""; for(int __iii__ = min(x1,SIZE(b)),  __jjj__ = min(x2, SIZE(b)-1); __iii__ <= __jjj__; ++__iii__){ t += b[__iii__] + '0'; } return '"' + t + '"'; } template <typename A, typename... C> string to_string(const A (&v), int x1=0, int x2=1e9, C... coords); int l_v_l_v_l = 0, t_a_b_s = 0; template <typename A, typename B> string to_string(const pair<A, B> &p) { l_v_l_v_l++; string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")"; l_v_l_v_l--; return res; } template <typename A, typename... C> string to_string(const A (&v), int x1, int x2, C... coords) { int rnk = rank<A>::value; string tab(t_a_b_s, ' '); string res = ""; bool first = true; if(l_v_l_v_l == 0) res += n_l; res += tab + "["; x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v)); auto l = begin(v); advance(l, x1); auto r = l; advance(r, (x2-x1) + (x2 < SIZE(v))); for (auto e = l; e != r; e = next(e)) { if (!first) { res += ", "; } first = false; l_v_l_v_l++; if(e != l){ if(rnk > 1) { res += n_l; t_a_b_s = l_v_l_v_l; }; } else{ t_a_b_s = 0; } res += to_string(*e, coords...); l_v_l_v_l--; } res += "]"; if(l_v_l_v_l == 0) res += n_l; return res; } void debugm(){;} template<typename Heads, typename... Tails> void debugm(Heads H, Tails... T){ cerr << to_string(H) << " | "; debugm(T...); } 
#define debugm(...) cerr << "[" << #__VA_ARGS__ << "]: "; debugm(__VA_ARGS__); cerr << endl

double round(double num, int dec) {return std::round(num*pow(10,dec))/pow(10,dec);}
vector<string> split(string s, char delim) {replace(s.begin(),s.end(),delim,' ');stringstream ss(s); string buf;vector<string> v;while (ss>>buf) v.push_back(buf);return v;}
template<typename T, int size> int siz(T(&)[size]){return size;}
long long POW(long long base, long long exp) {long long res = 1;while (exp > 0) {if (exp % 2 == 1){res= (res * base) % MOD;}exp = exp >> 1;base = (base * base) % MOD;}return res;}



int n, m, k;
vi friends[100005];
vi blocks[100005];
bool visited[100005];
int idx[100005];
set<int> group[100005];
int group_id = 1;

void dfs(int node) {
    visited[node] = true;
    idx[node] = group_id;
    group[group_id].insert(node);
    for (int child : friends[node])
        if (!visited[child]) dfs(child);
}

int main() {
    cin >> n >> m >> k;
    rep(i, m) {
        int u, v; cin >> u >> v;
        friends[u].pb(v);
        friends[v].pb(u);
    }
    rep(i, k) {
        int u, v; cin >> u >> v;
        blocks[u].pb(v);
        blocks[v].pb(u);
    }

    repn(i, 1, n+1) {
        if (!visited[i]) {
            dfs(i);
            group_id++;
        }
    }

    repn(i, 1, n+1) {
        int group_idx = idx[i];
        int ans = group[group_idx].size() - 1 - friends[i].size();
        for (int bad : blocks[i]) {
            if (group[group_idx].count(bad)) ans--;
        }
        cout << ans << " ";
    }
}