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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Collections;
using System.Collections.Specialized;
using System.Linq;
using System.Text;
using System.IO;
using System.Reflection;
using static System.Math;
using System.Numerics;
static class Program{
	const int mod=(int)1e9+7;
	static void Main(){
		Sc sc=new Sc();
		var s=sc.Ia;
		Uf uf=new Uf(s[0]+1);
		var h=new int[s[0]+1];
		var b=new int[s[0]+1];
		var ans=new int[s[0]];
		for(int i = 0;i<s[1];i++) {
			var e=sc.Ia;
			uf.Union(e[0],e[1]);
			h[e[0]]++;
			h[e[1]]++;
		}
		for(int i = 0;i<s[2];i++) {
			var e=sc.Ia;
			if(uf.Same(e[0],e[1])){
				b[e[0]]++;
				b[e[1]]++;
			}
		}
		for(int i = 1;i<=s[0];i++) {
			Console.WriteLine("{0} {1} {2}",uf.Size(i),h[i],b[i]);
			ans[i-1]=uf.Size(i)-h[i]-b[i]-1;
		}
		Console.WriteLine("{0}",String.Join(" ",ans));
	}
}
public class Uf{
	private int[] arr,hi;
	public int all;
	public Uf(int n){
		arr=new int[n];
		hi=new int[n];
		all=n;
		for(int i=0;i<n;i++){arr[i]=-1;}
	}
	public void Union(int a,int b){
		a=Root(a);
		b=Root(b);
		if(a!=b){
			if(hi[a]<hi[b]){a^=b;b^=a;a^=b;}
			arr[a]+=arr[b];
			hi[a]=Math.Max(hi[a],hi[b]+1);
			arr[b]=a;
			all--;
		}
	}
	public bool Same(int a,int b){return Root(a)==Root(b);}
	public int Size(int a){return -arr[Root(a)];}
	public int Root(int a){return arr[a]<0?a:arr[a]=Root(arr[a]);}
}
public class Sc{
	public int I{get{return int.Parse(Console.ReadLine());}}
	public long L{get{return long.Parse(Console.ReadLine());}}
	public double D{get{return double.Parse(Console.ReadLine());}}
	public string S{get{return Console.ReadLine();}}
	public int[] Ia{get{return Array.ConvertAll(Console.ReadLine().Split(),int.Parse);}}
	public long[] La{get{return Array.ConvertAll(Console.ReadLine().Split(),long.Parse);}}
	public double[] Da{get{return Array.ConvertAll(Console.ReadLine().Split(),double.Parse);}}
	public string[] Sa{get{return Console.ReadLine().Split();}}
	public object[] Oa{get{return Console.ReadLine().Split();}}
	public int[] Ia2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),int.Parse);}}
	public int[] Ia3(int a){return Array.ConvertAll((a.ToString()+" "+Console.ReadLine()).Split(),int.Parse);}
	public int[] Ia3(bool a,int b,bool c,int d){return Array.ConvertAll(((a?b.ToString()+" ":"")+Console.ReadLine()+(c?" "+d.ToString():"")).Split(),int.Parse);}
	public long[] La2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),long.Parse);}}
	public long[] La3(int a){return Array.ConvertAll((a.ToString()+" "+Console.ReadLine()).Split(),long.Parse);}
	public long[] La3(bool a,int b,bool c,int d){return Array.ConvertAll(((a?b.ToString()+" ":"")+Console.ReadLine()+(c?" "+d.ToString():"")).Split(),long.Parse);}
	public double[] Da2{get{return Array.ConvertAll(("0 "+Console.ReadLine()+" 0").Split(),double.Parse);}}
	public double[] Da3(int a){return Array.ConvertAll((a.ToString()+" "+Console.ReadLine()).Split(),double.Parse);}
	public double[] Da3(bool a,int b,bool c,int d){return Array.ConvertAll(((a?b.ToString()+" ":"")+Console.ReadLine()+(c?" "+d.ToString():"")).Split(),double.Parse);}
	public T[] Arr<T>(int n,Func<T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f();}return a;}
	public T[] Arr<T>(int n,Func<int,T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(i);}return a;}
	public T[] Arr<T>(int n,Func<string[],T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(Console.ReadLine().Split());}return a;}
	public T[] Arr<T>(int n,Func<int,string[],T> f){var a=new T[n];for(int i=0;i<n;i++){a[i]=f(i,Console.ReadLine().Split());}return a;}
}