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 int long long
#define rep(i, n) for (int i = (int)(0); i < (int)(n); ++i)
#define reps(i, n) for (int i = (int)(1); i <= (int)(n); ++i)
#define rrep(i, n) for (int i = ((int)(n)-1); i >= 0; i--)
#define rreps(i, n) for (int i = ((int)(n)); i > 0; i--)
#define irep(i, m, n) for (int i = (int)(m); i < (int)(n); ++i)
#define ireps(i, m, n) for (int i = (int)(m); i <= (int)(n); ++i)
#define SORT(v, n) sort(v, v + n);
#define vsort(v) sort(v.begin(), v.end());
#define all(v) v.begin(), v.end()
#define mp(n, m) make_pair(n, m);
#define cout(d) cout<<d<<endl;
#define coutd(d) cout<<std::setprecision(10)<<d<<endl;
#define cinline(n) getline(cin,n);
#define replace_all(s, b, a) replace(s.begin(),s.end(), b, a);
#define PI (acos(-1))
#define FILL(v, n, x) fill(v, v + n, x);
#define sz(x) int(x.size())
#define pqasc priority_queue<long long, vector<long long>, greater<long long>>

using ll = long long;
using vi = vector<int>;
using vvi = vector<vi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using pii = pair<int, int>;
using pll = pair<ll, ll>;

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; }

class UnionFind{
private:
  vector<int> Parent; 
 public:
  UnionFind(int N){
    Parent = vector<int>(N, -1);
  }
  
  int root(int A){
    if(Parent[A] < 0) return A;
    return Parent[A] = root(Parent[A]);
  }
  
  int size(int A){
    return -Parent[root(A)]; 
  }
  
  bool connect(int A, int B){
    A = root(A);
    B = root(B);
    if(A == B){
      return false;
    }
    if(size(A) < size(B)) swap(A, B);
    Parent[A] += Parent[B];
    Parent[B] = A;
    return true;
  }
  
  bool isSame(int A, int B){
    return root(A) == root(B);
  }
};

long long gcd(long long a, long long b) { return b ? gcd(b, a%b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }

long long ngcd(vector<long long> a) {
  long long res = a[0];
  for(int i=1; i<a.size() && res!=1; i++) res = gcd(a[i], res);
  return res;
}

long long nlcm(vector<long long> a){
  long long res = a[0];
  for(int i=1; i<a.size(); i++) res = lcm(a[i], res);
  return res;
}

bool is_prime(long long n) {
  if(n < 2) return false;
  for (long long i = 2; i * i <= n; i++) if (n % i == 0) return false;
  return true;
}

int digsum(int n) {
  int res=0;
  while(n) res += n%10, n /= 10;
  return res;
}

int digcnt(int n){
  int res=0;
  while(n) res++, n /= 10;
  return res;
}

vector<int> divisor(int n) {
  vector<int> res;
  for (int i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      res.push_back(i);
      if (i != n / i) res.push_back(n / i);
    }
  }
  return res;
}

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) %= mod;
    n >>= 1;
  }
  return res;
}

long long comb(long long a, long long b, long long mod){
  long long res = 1;
  for (long long i = a; i > a - b; i--) res = res * i % mod;
  for (long long i = 1; i <= b; i++) res = (res * modpow(i, mod-2, mod)) % mod;
  return res;
}

const int dy[] = {0, 1, 0, -1, -1, 1, 1, -1};
const int dx[] = {1, 0, -1, 0, 1, 1, -1, -1};

inline bool inside(int y, int x, int H, int W) {
    return (y >= 0 && x >= 0 && y < H && x < W);
}

const int INF = 1e9;
const int MOD = 998244353;
const ll LINF = 1e18;

long long fact(int a){
  long long res=1;
  for(int i=1; i<=a; i++) (res *= i) %= MOD;
  return res;
}

void add(long long& a, long long b){
  a = (a+b) % MOD;
}

void mul(long long& a, long long b){
  a = (a*b) % MOD;
}

char G[10][10];

signed main()
{
  cin.tie( 0 ); ios::sync_with_stdio( false );
  int n,m,k; cin>>n>>m>>k;
  UnionFind uf(n);
  map<int,int> x,y;
  map<pii,int> z;
  rep(i,m){
    int a,b; cin>>a>>b; a--,b--;
    x[a]++;
    x[b]++;
    z[pii(a,b)]=z[pii(b,a)]=1;
    uf.connect(a,b);
  }
  rep(i,k){
    int c,d; cin>>c>>d; c--,d--;
    if(!uf.isSame(c,d)) continue;
    if(z[pii(c,d)]) continue;
    y[c]++; y[d]++;
  }
  
  rep(i,n){
    cout<<uf.size(i)-1-x[i]-y[i]<<" ";
  }
  cout<<endl;
}