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;
ll mp[1000010],r[1000010],sz[1000010];
ll get_root(ll x){
	if(mp[x]==x) return x;
	else mp[x]=get_root(mp[x]);
	return mp[x];
}
void combine(ll x,ll y){
	if(r[x]<r[y]){
        sz[get_root(y)]+=sz[get_root(x)];
        mp[get_root(x)]=get_root(y);

    }
	else if(r[y]<r[x]){
        sz[get_root(x)]+=sz[get_root(y)];
        mp[get_root(y)]=get_root(x);
    }
	else {
        sz[get_root(y)]+=sz[get_root(x)];
		mp[get_root(x)]=get_root(y);
		r[get_root(y)]+=1;
	}
}
bool check(ll x,ll y){
	return get_root(x)==get_root(y);
}
void init(){
    
    for (ll i=0;i<1000010;i++){
		mp[i]=i;
        sz[i]=1;
	}
}
int main() {
	ios_base::sync_with_stdio(false);
    cin.tie(0);
	ll n,e,i,x,y,sum=0,k,m;
	cin>>n>>m>>k;
    ll outdegree[n+1];
    for (i=1;i<=n;i++)outdegree[i]=0;
	init();
	for (i=0;i<m;i++){
		cin>>x>>y;
        outdegree[x]++;
        outdegree[y]++;
		if (!check(x,y))combine(x,y);
	}
    ll ans[n+1];
	for (i=1;i<=n;i++){
		ans[i]=sz[get_root(i)]-1-outdegree[i];
	}
    sum/=2;
    for (i=0;i<k;i++){
        cin>>x>>y;
        if (check(x,y)){
            ans[x]--;
            ans[y]--;
        }
    }
	for (i=1;i<=n;i++){
        if (i!=1)cout<<' ';
        cout<<ans[i];
    }
    
}