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>
#define _GLIBCXX_DEBUG
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define all(v) v.begin(), v.end()
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; }
using ll = long long;
const ll MOD = 1e9+7; int INF=INT_MAX; ll LLINF = 1LL << 60;
const vector<int> d8x = {1, 0, -1, 0, 1, -1, -1, 1}; vector<int> d8y = {0, 1, 0, -1, 1, 1, -1, -1};
const vector<int> d4x = {1, 0, -1, 0}; vector<int> d4y = {0, 1, 0, -1};

//

struct union_find {
    vector<int> par;

    union_find(int n) : par(n,-1) {}

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

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

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

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

int main(){
    int n,m,k;
    cin>>n>>m>>k;
    vector<set<int>> dame(n);
    union_find uf(n);

    for(int i=0; i<m; ++i){
        int a,b;
        cin>>a>>b;
        --a;
        --b;
        dame[a].insert(b);
        dame[b].insert(a);
        uf.unite(a,b);
    }

    for(int i=0; i<k; ++i){        
        int c,d;
        cin>>c>>d;
        --c;
        --d;
        if(!uf.same(c,d)) continue;
        dame[c].insert(d);
        dame[d].insert(c);
    }

    for(int i=0; i<n; ++i){
        int mem=uf.size(i) -1;
        mem-=dame[i].size();
        cout<<mem<<" ";
    }
    cout<<endl;
}