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;
typedef long long ll;
#define MOD 1000000007
#define INF 1e9
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()

struct UnionFind{
    vector<int> par;

    UnionFind(int n) : par(n,-1) { }
    void init(int n) { par.assign(n,-1); }

    int root(int x){
        if(par[x]<0) return x;
        else return par[x]=root(par[x]);
    }

    bool issame(int x,int y){
        return root(x)==root(y);
    }

    bool merge(int x,int y){
        x=root(x);
        y=root(y);
        if(x==y) return false;
        if(par[x]>par[y]) swap(x,y);
        par[x]+=par[y];
        par[y]=x;
        return true;
    }

    int size(int x){
        return -par[root(x)];
    }
};

int main() {
    int n,m,k;
    cin>>n>>m>>k;

    vector<vector<int>> G(n);
    UnionFind uf(n);
    rep(i,m){
        int a,b;
        cin>>a>>b;
        a--,b--;
        uf.merge(a,b);
        G[a].push_back(b);
        G[b].push_back(a);
    }

    rep(i,k){
        int c,d;
        cin>>c>>d;
        c--,d--;
        if(uf.issame(c,d)){
            G[c].push_back(d);
            G[d].push_back(c);
        }
    }

    rep(i,n){
        cout<<uf.size(i)-1-G[i].size()<<" ";
    }
    cout<<endl;

}