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: Wrong Answer)

#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
#include<stdio.h>
#include<vector>
#include<queue>
#include<math.h>
using namespace std;
#define int long long
#define double long double
#define rep(s,i,n) for(int i=s;i<n;i++)
#define c(n) cout<<n<<endl;
#define ic(n) int n;cin>>n;
#define sc(s) string s;cin>>s;
#define mod 1000000007
#define inf 1000000000000000007
#define f first
#define s second
#define mini(c,a,b) *min_element(c+a,c+b)
#define maxi(c,a,b) *max_element(c+a,c+b)
#define pi 3.141592653589793238462643383279
#define e_ 2.718281828459045235360287471352
#define P pair<int,int>
#define upp(a,n,x) upper_bound(a,a+n,x)-a;
#define low(a,n,x) lower_bound(a,a+n,x)-a;
#define UF UnionFind 
//printf("%.12Lf\n",);
int keta(int x){
	rep(0,i,30){
		if(x<10){
			return i+1;
		}
		x=x/10;
	}
}	
int gcd(int x,int y){
	int aa=x,bb=y;
	rep(0,i,1000){
		aa=aa%bb;
		if(aa==0){
			return bb;
		}	
		bb=bb%aa;
		if(bb==0){
			return aa;
		}
	}
}	
int lcm(int x,int y){
	int aa=x,bb=y;
	rep(0,i,1000){
		aa=aa%bb;
		if(aa==0){
			return x/bb*y;
		}	
		bb=bb%aa;
		if(bb==0){
			return x/aa*y;
		}
	}
}
bool p(int x){
	if(x==1)return false;
	rep(2,i,sqrt(x)+1){
		if(x%i==0&&x!=i){
			return false;
		}
	}	
	return true;
}
int max(int a,int b){
	if(a>=b)return a;
	else return b;
}
string maxst(string s,string t){
	int n=s.size();
	int m=t.size();
	if(n>m)return s;
	else if(n<m)return t;
	else{
		rep(0,i,n){
			if(s[i]>t[i])return s;
			if(s[i]<t[i])return t;
		}
		return s;
	}
}		
int min(int a,int b){
	if(a>=b)return b;
	else return a;
}	
int n2[41];		
int nis[41]; 
int nia[41];
int mody[41];
int nn;
int com(int n,int y){
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	int bunsi=1,bunbo=1;
	rep(0,i,y)bunsi=(bunsi*(n-i))%mod;
	rep(0,i,y)bunbo=(bunbo*(i+1))%mod;
	mody[0]=bunbo;
	rep(1,i,41){
		bunbo=(bunbo*bunbo)%mod;
		mody[i]=bunbo;
	}
	rep(0,i,41)nis[i]=0;
	nn=mod-2;
	for(int i=40;i>=0;i-=1){
		if(nn>n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	nis[0]++;
	rep(0,i,41){
		if(nis[i]==1){
			bunsi=(bunsi*mody[i])%mod;
		}
	}
	return bunsi;
}
int gyakugen(int n,int y){
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	mody[0]=y;
	rep(1,i,41){
		y=(y*y)%mod;
		mody[i]=y;
	}
	rep(0,i,41)nis[i]=0;
	nn=mod-2;
	for(int i=40;i>=0;i-=1){
		if(nn>n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	nis[0]++;
	rep(0,i,41){
		if(nis[i]==1){
			n=(n*mody[i])%mod;
		}
	}
	return n;	
}	
int yakuwa(int n){
	int sum=0;
	rep(1,i,sqrt(n+1)){
		if(n%i==0)sum+=i+n/i;
		if(i*i==n)sum-=i;
	}
	return sum;
}	
bool integer(double n){
	if(n==long(n))return true;
	else return false;	
}
int poow(int y,int n){	
	if(n==0)return 1;
	n-=1;
	int ni=1;
	for(int i=0;i<41;i++){
		n2[i]=ni;
		ni*=2;
	}
	int yy=y;
	mody[0]=yy;
	rep(1,i,41){
		yy=(yy*yy)%mod;
		mody[i]=yy;
	}
	rep(0,i,41)nis[i]=0;
	nn=n;
	for(int i=40;i>=0;i-=1){
		if(nn>=n2[i]){
			nis[i]++;
			nn-=n2[i];
		}
	}
	rep(0,i,41){
		if(nis[i]==1){
			y=(y*mody[i])%mod;
		}
	}
	return y;	
}
int minpow(int x,int y){
	int sum=1;
	rep(0,i,y)sum*=x;
	return sum;
}
int ketawa(int x,int sinsuu){
	int sum=0;
	rep(0,i,100)sum+=(x%poow(sinsuu,i+1))/(poow(sinsuu,i));
	return sum;
}	
int sankaku(int a){
	return a*(a+1)/2;
} 
int same(int a[1111111],int n){
	int ans=0;
	rep(0,i,n){
		if(a[i]==a[i+1]){
			int j=i;
			while(a[j+1]==a[i]&&j<=n-2)j++;
			ans+=sankaku(j-i);
			i=j;
		}
	}
	return ans;
}
bool search(int x){
	int n;
	int a[n];
	int l=0,r=n;
	while(r-l>=1){
		int i=(l+r)/2;
		if(a[i]==x)return true;
		else if(a[i]<x)l=i+1;
		else r=i;
	}
	return false;
}
/*using Graph=vector<vector<int>>;
int oya[114514];
int depth[114514];
void dfs(const Graph &G, int v, int p, int d) {
    depth[v] = d;
    oya[v]=p;
    for (auto nv : G[v]) {
       if (nv == p) continue; // nv が親 p だったらダメ
       dfs(G, nv, v, d+1); // d を 1 増やして子ノードへ
    }
}*/
/*int H=10,W=10;
char field[10][10];
char memo[10][10];
void dfs(int h, int w) {
    memo[h][w] = 'x';

    // 八方向を探索
    for (int dh = -1; dh <= 1; ++dh) {
        for (int dw = -1; dw <= 1; ++dw) {
        	if(abs(0-dh)+abs(0-dw)==2)continue;
            int nh = h + dh, nw = w + dw;

            // 場外アウトしたり、0 だったりはスルー
            if (nh < 0 || nh >= H || nw < 0 || nw >= W) continue;
            if (memo[nh][nw] == 'x') continue;

            // 再帰的に探索
            dfs(nh, nw);
        }
    }
}*/
int XOR(int a,int b){
	int ni=1;
	rep(0,i,41){
		n2[i]=ni;
		ni*=2;
	}
	rep(0,i,41)nis[i]=0;
	for(int i=40;i>=0;i-=1){
		if(a>=n2[i]){
			nis[i]++;
			a-=n2[i];
		}
		if(b>=n2[i]){
			nis[i]++;
			b-=n2[i];
		}	
	}
	int sum=0;
	rep(0,i,41)sum+=(nis[i]%2*n2[i]);
	return sum;
}
//int ma[1024577][21];
//for(int bit=0;bit<(1<<n);bit++)rep(0,i,n)if(bit&(1<<i))ma[bit][i]=1;
struct UnionFind {
    vector<int> par; // par[i]:iの親の番号　(例) par[3] = 2 : 3の親が2

    UnionFind(int N) : par(N) { //最初は全てが根であるとして初期化
        for(int i = 0; i < N; i++) par[i] = i;
    }

    int root(int x) { // データxが属する木の根を再帰で得る：root(x) = {xの木の根}
        if (par[x] == x) return x;
        return par[x] = root(par[x]);
    }

    void unite(int x, int y) { // xとyの木を併合
        int rx = root(x); //xの根をrx
        int ry = root(y); //yの根をry
        if (rx == ry) return; //xとyの根が同じ(=同じ木にある)時はそのまま
        par[rx] = ry; //xとyの根が同じでない(=同じ木にない)時：xの根rxをyの根ryにつける
    }

    bool same(int x, int y) { // 2つのデータx, yが属する木が同じならtrueを返す
        int rx = root(x);
        int ry = root(y);
        return rx == ry;
    }
};
P fri[114514];
P blo[114514];
int cnt[114514];
P ne[114514];
int ans[114514];
signed main(){
	ic(n) ic(m) ic(k)
	UF tree(n);
	rep(0,i,m){
		cin>>fri[i].f>>fri[i].s;
		fri[i].f-=1;
		fri[i].s-=1;
		tree.unite(fri[i].f,fri[i].s);
		cnt[fri[i].f]++;
		cnt[fri[i].s]++;
	}	
	rep(0,i,k){
		cin>>blo[i].f>>blo[i].s;
		blo[i].f-=1;
		blo[i].s-=1;
		cnt[blo[i].f]++;
		cnt[blo[i].s]++;
	}
	rep(0,i,n){
		ne[i].f=tree.root(i);
		ne[i].s=i;
	}
	sort(ne,ne+n);
	rep(0,i,n-1){
		if(ne[i].f==ne[i+1].f){
			int j=i;
			while(ne[j+1].f==ne[i].f)j++;
			rep(i,l,j+1){
				ans[ne[l].s]+=j-i+1;
			}
			i=j;
		}
	}	
	rep(0,i,m){
		if(tree.same(fri[i].f,fri[i].s)){
			ans[fri[i].f]-=1;
			ans[fri[i].s]-=1;
		}
	}
	rep(0,i,k){
		if(tree.same(blo[i].f,blo[i].s)){
			ans[blo[i].f]-=1;
			ans[blo[i].s]-=1;
		}
	}
	rep(0,i,n)cout<<ans[i]-1<<" ";	
}