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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
using vi = vector<int>;
using vvi = vector<vi>;
#define ote(x) cout<<(x)<<endl
#define rp(i,s,e) for(int i=(s);i<(e);++i)
int N,M,K;

#define TMAX 100000
vvi F,B;//friend,block
// vi visited(TMAX+1);
vi gid(TMAX+1);
vi ans(TMAX+1);

// void dfs(int root, int parent){
// 	if(visited[parent]) return;
// 	visited[parent] = true;
// 	gid[root]++;
// 	for(auto c:F[parent]){
// 		dfs(root, c);
// 	}
// 	visited[parent] = false;
// }

void bfs(int root){
	vi visited(N+1);
	queue<int> q;
	q.push(root);
	while(!q.empty()){
		int p = q.front(); q.pop();
		// cout<<p<<"\n";//
		visited[p] = true;
		gid[p] = root;
		for(auto c:F[p]){
			if(visited[c]) continue;
			q.push(c);
		}
	}
}


int block_calc(int i){
	int ret = B[i].size();
	for(auto b:B[i]){
		if(gid[b] != gid[i]) ret--;
	}
	return ret;
}

void solve(){
	rp(i,1,N+1){
		bfs(i);
		// dfs(i, i);
	}
	// rp(i,1,N+1){printf("%2d ",i);}puts("");///
	// rp(i,1,N+1){printf("%2d ",gid[i]);}puts("");///
	vi cnt(N+1);
	rp(i,1,N+1){
		cnt[gid[i]]++;
	}
	int myself = 1;
	// rp(i,1,N+1){printf("%2d ",cnt[i]);}puts("");///
	rp(i,1,N+1){
		// ans[i] = cnt[gid[i]] - F[i].size() - B[i].size() - 1;
		// cout<<i<<" "<<block_calc(i)<<"\n";//
		ans[i] = cnt[gid[i]] - F[i].size() - block_calc(i) - myself;
	}
	// rp(i,1,N+1)printf("%2d " ,ans[i]);puts("");///

}

int main(){
	cin>>N>>M>>K;

	F.resize(N+1);
	rp(i,0,M){
		int s,t; cin>>s>>t;
		F[s].push_back(t);
		F[t].push_back(s);
	}

	B.resize(N+1);
	rp(i,0,K){
		int s,t; cin>>s>>t;
		B[s].push_back(t);
		B[t].push_back(s);
	}

	// rp(i,0,N+1){
	// 	cout<<i<<") ";
	// 	for(auto x:F[i]){
	// 		cout<<x<<" ";
	// 	}
	// 	puts("");
	// }
	// puts("");

	// rp(i,0,N+1){
	// 	cout<<i<<") ";
	// 	for(auto x:B[i]){
	// 		cout<<x<<" ";
	// 	}
	// 	puts("");
	// }
	// puts("");

	solve();
	rp(i,1,N+1)cout<<ans[i]<<" ";
}