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;
using ll = long long;
using P = pair<int, int>;
#define rep(i, n) for(int (i)=0;(i)<(n);(i)++)
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
bool custom1 (pair<int, int> a, pair<int, int> b) {
  return (a.first < b.first);
}
bool custom2(pair<int, int> a, pair<int, int> b) {
  return (a.second < b.second);
}

const int MAX = 200005;
const long long MOD = 1000000007;
const long long MODMOD = 998244353;
const double PI = acos(-1);

long long fac[MAX], finv[MAX], inv[MAX];

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

long long modpow(long long a, long long n, long long mod) {
    long long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}
struct UnionFind {
  vector<int> p;
  vector<int> s;
  void init(int n) {
    p.resize(n, -1);
    s.resize(n, 1);
  }
  int find(int n) {
    if(p[n] == -1) return n;
    else return find(p[n]);
  }
  int size(int n) {
    return s[find(n)];
  }
  void unite(int n, int m) {
    int pn = find(n);
    int pm = find(m);
    if(pn == pm) return;
    if(s[pn] > s[pm]) {
      s[pn] = s[pn] + s[pm];
      p[pm]= pn;
    } else {
      s[pm] = s[pm] + s[pn];
      p[pn] = pm;
    }
  }
};

long long gcd(long long a, long long b) {
  if(a % b == 0) return b;
  else return gcd(b, a % b);
}
int main() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> friends(n);
  UnionFind uf;
  uf.init(n);
  rep(i, m) {
    int a, b;
    cin >> a >> b;a--;b--;
    friends[a]++;friends[b]++;
    uf.unite(a, b);
  }
  vector<int> candidate(n);
  rep(i, n) {
    candidate[i] = uf.size(i) - friends[i] - 1;
  }
  rep(i, k) {
    int c,d;
    cin >> c >> d;c--;d--;
    if(uf.find(c) == uf.find(d)) {
      candidate[c]--;candidate[d]--;
    }
  }
  rep(i, n) {
    cout << candidate[i];
    if(i == n-1) cout << endl;
    else cout << " ";
  }
  return 0;
}