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: Runtime Error)

#include<iostream>
#include <cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<iomanip>
#include<math.h>
#include<complex>
#include<queue>
#include<deque>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<functional>
#include<assert.h>
#include<numeric>
using namespace std;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
using ll = long long;
typedef pair<int,int> pint;
typedef pair<ll,int> pli;
const int inf=1e9+7;
const ll longinf=1LL<<60 ;
const ll mod=1e9+7 ;

struct UnionFind{
	vector<int> par,siz;
	UnionFind(int N): par(N), siz(N){
		rep(i, N){
			par[i] = i; siz[i] = 1;
		}
	}
	int root(int x){
		if(par[x]==x) return x;
		else return root(par[x]);
	}
	void unite(int x, int y){
		x=root(x); y=root(y);
		if(x==y) return;
		if(siz[x] < siz[y]) swap(x, y);
		par[y] = x;
		siz[x] += siz[y];
	}
	bool same(int x, int y){
		return root(x)==root(y);
	}
	int size(int x){
		return siz[root(x)];
	}
};

int main(){
  int N, M, K; cin >> N >> M >> K;
  UnionFind uf(N+100); bool G[N+100][N+100] = {}, B[N+100][N+100] = {};
  vector<int> to[N+100]; int mai[N+100] = {};
  rep(i, M){
    int a, b; cin >> a >> b; a--; b--;
    uf.unite(a, b);
    G[a][b] = true;
    to[a].push_back(b); to[b].push_back(a);
  }
  rep(i, K){
    int c, d; cin >> c >> d;
    c--; d--;
    B[c][d]=true;
    if(uf.same(c, d) && !G[c][d]){
      mai[c]++; mai[d]++;
    }
  }
  rep(i, N){
    int ans = uf.size(i);
    ans-=to[i].size();
    ans--;
    ans-=mai[i];
    cout << ans;
    if(i!=N-1) cout << " ";
  }
  cout << endl;
}