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 <iostream>
#include <utility>
#include <vector>
#include <string>
#define rep(i, n)   for(int i = 0; i < (n); i++)
using namespace std;
typedef long long ll;

//集合を表すデータ構造、根は大きさを負で持ち、子は根を正で持つ
//[0]に全体の集合の数を入れる
//swapを使用するため<utility>をインクルードする
struct UnionFind{
    vector<int>data;
    UnionFind(int size) :   data(size+1, -1){
        data[0] = size;
    }

    //集合のmerge
    //すでに同じならfalse
    bool merge(int x, int y){
        x = root(x);    y = root(y);
        if(x == y)  return false;
        //要素数の少ない方を多い方に繋げる
        if(data[y] < data[x])   swap(x, y);
        data[x] += data[y];
        data[y] = x;
        data[0]--;
        return true;
    }

    //どの集合に属しているか
    int root(int x){
        //根に繋ぎ直す
        return (data[x] < 0) ? x : (data[x] = root(data[x]));
    }

    //ある要素が属する集合の大きさ
    int size(int x){
        return -data[root(x)];
    }

    //集合の数
    int set(){
        return data[0];
    }
};

int main(){
    int n,m,k;  cin >> n >> m >> k;
    UnionFind frend(n);
    vector<vector<int> >F(n+1);
    vector<vector<int> >B(n+1);
    rep(i, m){
        int a,b;    cin >> a >> b;
        frend.merge(a, b);
        F[a].push_back(b);
        F[b].push_back(a);
    }
    rep(i, k){
        int c,d;    cin >> c >> d;
        B[c].push_back(d);
        B[d].push_back(c);
    }
    vector<int>ans(n);
    rep(i, n){
        int now = i+1;
        int tmp = frend.size(now);
        tmp--;  //自身を引く
        tmp -= F[now].size();
        rep(j, B[now].size()){
            if(frend.root(now) == frend.root(B[now][j]))    tmp--;
        }
        ans[i] = tmp;
    }
    rep(i, n){
        printf("%d%c", ans[i], (i == n-1)?'\n':' ');
    }
}