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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll=long long;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vl = vector<ll>;
using vvl = vector<vector<ll>>;
#define all(x) x.begin(),x.end()
#define rep(i,j,n) for (long long i = j; i < (long long)(n); i++)
#define _GLIBCXX_DEBUG
#define MOD 1000000007
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; }
//(a+b-1)/b
//priority_queue<ll, vector<ll>, greater<ll>> q;
using pl = pair<ll,ll>;

class UnionFind
{
private:
    vector<ll> p;
    vector<ll> rank;
public:
    UnionFind(ll n){
        p.resize(n,-1);
        rank.resize(n,1);
    }
    ll find(ll x){
        if(p[x]<0)return x;
        return p[x]=find(p[x]);
    }
    void unite(ll x,ll y){
        x=find(x);
        y=find(y);
		if(x==y)return;
        if(rank[x]>rank[y])swap(x,y);
        if(rank[x]==rank[y])rank[y]++;
        p[y]+=p[x];
        p[x]=y;
    }
    ll size(ll x){
        return -p[find(x)];
    }
};



signed main(){
	ll n, m, k; cin >> n >> m >> k;
	vl fr(n);
	UnionFind uf(n);
	rep(i,0,m){
		ll a,b; cin >> a >> b;
		--a; --b;
		fr[a]++;
		fr[b]++;
		uf.unite(a,b);
	}
	vl to[n];
	rep(i,0,k){
		ll a,b; cin >> a >> b;
		--a; --b;
		to[a].push_back(b);
		to[b].push_back(a);
	}
	rep(i,0,n){
		ll ans=uf.size(i)-fr[i]-1;
		for(ll c:to[i]){
			if(uf.find(c)==uf.find(i))--ans;
		}
		cout<<ans<<" ";
	}
	cout<<endl;
	

	return 0;
}