Problem: Alice and Bob are going to celebrate Christmas by playing a game with a tree of presents. The tree has n nodes (numbered 1 to n, with some node r as its root). There are a_i presents are hanging from the i-th node.

Before beginning the game, a special integer k is chosen. The game proceeds as follows:

  * Alice begins the game, with moves alternating each turn;
  * in any move, the current player may choose some node (for example, i) which has depth at least k. Then, the player picks some positive number of presents hanging from that node, let's call it m (1 ≤ m ≤ a_i);
  * the player then places these m presents on the k-th ancestor (let's call it j) of the i-th node (the k-th ancestor of vertex i is a vertex j such that i is a descendant of j, and the difference between the depth of j and the depth of i is exactly k). Now, the number of presents of the i-th node (a_i) is decreased by m, and, correspondingly, a_j is increased by m;
  * Alice and Bob both play optimally. The player unable to make a move loses the game.



For each possible root of the tree, find who among Alice or Bob wins the game.

Note: The depth of a node i in a tree with root r is defined as the number of edges on the simple path from node r to node i. The depth of root r itself is zero.

Input

The first line contains two space-separated integers n and k (3 ≤ n ≤ 10^5, 1 ≤ k ≤ 20).

The next n-1 lines each contain two integers x and y (1 ≤ x, y ≤ n, x ≠ y), denoting an undirected edge between the two nodes x and y. These edges form a tree of n nodes.

The next line contains n space-separated integers denoting the array a (0 ≤ a_i ≤ 10^9).

Output

Output n integers, where the i-th integer is 1 if Alice wins the game when the tree is rooted at node i, or 0 otherwise.

Example

Input


5 1
1 2
1 3
5 2
4 3
0 3 2 4 4


Output


1 0 0 1 1 

Note

Let us calculate the answer for sample input with root node as 1 and as 2.

Root node 1

Alice always wins in this case. One possible gameplay between Alice and Bob is:

  * Alice moves one present from node 4 to node 3. 
  * Bob moves four presents from node 5 to node 2. 
  * Alice moves four presents from node 2 to node 1. 
  * Bob moves three presents from node 2 to node 1. 
  * Alice moves three presents from node 3 to node 1. 
  * Bob moves three presents from node 4 to node 3. 
  * Alice moves three presents from node 3 to node 1. 



Bob is now unable to make a move and hence loses.

Root node 2

Bob always wins in this case. One such gameplay is:

  * Alice moves four presents from node 4 to node 3. 
  * Bob moves four presents from node 5 to node 2. 
  * Alice moves six presents from node 3 to node 1. 
  * Bob moves six presents from node 1 to node 2. 



Alice is now unable to make a move and hence loses.

Solution (in CPP):

Solution:
#include <iostream>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#define rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define reps(x, s) for(llint (x) = 0; (x) < (llint)(s).size(); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define sz(x) ((ll)(x).size())
#define ceil(x, y) (((x)+(y)-1) / (y))
#define all(x) (x).begin(),(x).end()
#define outl(...) dump_func(__VA_ARGS__)
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<ll, ll> P;

struct edge{
	ll to, cost;
	edge(){}
	edge(ll a, ll b){ to = a, cost = b;}
};
const ll dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1};

const ll mod = 1000000007;
//const ll mod = 998244353;

struct mint{
	ll x = 0;
	mint(ll y = 0){x = y; if(x < 0 || x >= mod) x = (x%mod+mod)%mod;}
	mint(const mint &ope) {x = ope.x;}
	
	mint operator-(){return mint(-x);}
	mint operator+(const mint &ope){return mint(x) += ope;}
	mint operator-(const mint &ope){return mint(x) -= ope;}
	mint operator*(const mint &ope){return mint(x) *= ope;}
	mint operator/(const mint &ope){return mint(x) /= ope;}
	mint& operator+=(const mint &ope){
		x += ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint &ope){
		x += mod - ope.x;
		if(x >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint &ope){
		x *= ope.x, x %= mod;
		return *this;
	}
	mint& operator/=(const mint &ope){
		ll n = mod-2; mint mul = ope;
		while(n){
			if(n & 1) *this *= mul;
			mul *= mul;
			n >>= 1;
		}
		return *this;
	}
	mint inverse(){return mint(1) / *this;}
	bool operator ==(const mint &ope){return x == ope.x;}
	bool operator !=(const mint &ope){return x != ope.x;}
};
mint modpow(mint a, ll n){
	if(n == 0) return mint(1);
	if(n % 2) return a * modpow(a, n-1);
	else return modpow(a*a, n/2);
}
istream& operator >>(istream &is, mint &ope){
	ll t; is >> t, ope.x = t;
	return is;
}
ostream& operator <<(ostream &os, mint &ope){return os << ope.x;}
ostream& operator <<(ostream &os, const mint &ope){return os << ope.x;}

bool exceed(ll x, ll y, ll m){return x >= m / y + 1;}
void mark(){ cout << "*" << endl; }
void yes(){ cout << "Yes" << endl; }
void no(){ cout << "No" << endl; }
ll gcd(ll a, ll b){if(b == 0) return a; return gcd(b, a%b);}


template<typename T>
ostream& operator << (ostream& os, vector<T>& vec) {
	for(int i = 0; i < vec.size(); i++) {
		os << vec[i] << (i + 1 == vec.size() ? "" : " ");
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, deque<T>& deq) {
	for(int i = 0; i < deq.size(); i++) {
		os << deq[i] << (i + 1 == deq.size() ? "" : " ");
	}
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, const pair<T, U>& pair_var) {
	os << "(" << pair_var.first << ", " << pair_var.second << ")";
	return os;
}
template<typename T, typename U>
ostream& operator << (ostream& os, map<T, U>& map_var) {
	for(typename map<T, U>::iterator itr = map_var.begin(); itr != map_var.end(); itr++) {
		os << "(" << itr->first << ", " << itr->second << ")";
		itr++;
		if(itr != map_var.end()) os << ",";
		itr--;
	}
	return os;
}
template<typename T>
ostream& operator << (ostream& os, set<T>& set_var) {
	for(typename set<T>::iterator itr = set_var.begin(); itr != set_var.end(); itr++) {
		os << *itr;
		++itr;
		if(itr != set_var.end()) os << " ";
		itr--;
	}
	return os;
}
template<typename T>
void outa(T a[], ll s, ll t)
{
	for(ll i = s; i <= t; i++){ cout << a[i]; if(i < t) cout << " ";}
	cout << endl;
}
void dump_func() {cout << endl;}
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
	cout << head;
	if(sizeof...(Tail) > 0) cout << " ";
	dump_func(std::move(tail)...);
}

typedef pair<ll, P> T;

ll n, k;
vector<ll> G[100005];
ll a[100005];
int parent[100005];
llint dp[100005][45], dp2[100005][45];

const llint e = 0; //
llint ope(llint a, llint b)
{
	return max(a, b); //
}

llint* get(int u, int v)
{
	if(parent[u] == v) return dp2[u];
	else return dp[v];
}

void dfs(int v, int p)
{
	parent[v] = p;
	for(int i = 0; i < G[v].size(); i++){
		if(G[v][i] == p) continue;
		dfs(G[v][i], v);
	}
	
	for(int i = 0; i < G[v].size(); i++){
		int u = G[v][i];
		if(u == p) continue;
		rep(j, 0, 2*k-1) dp[v][(j+1)%(2*k)] ^= get(v, u)[j];
	}
	dp[v][0] ^= a[v];
}

void dfs2(int v, int p)
{
	llint m = G[v].size();
	ll sum[45] = {};
	
	for(int i = 0; i < m; i++){
		int u = G[v][i];
		rep(j, 0, 2*k-1) sum[(j+1)%(2*k)] ^= get(v, u)[j];
	}
	
	for(int i = 0; i < m; i++){
		int u = G[v][i];
		if(u == p) continue;
		rep(j, 0, 2*k-1) dp2[u][j] = sum[j];
		rep(j, 0, 2*k-1) dp2[u][(j+1)%(2*k)] ^= get(v, u)[j];
		dp2[u][0] ^= a[v];
	}
	
	for(int i = 0; i < G[v].size(); i++){
		if(G[v][i] == p) continue;
		dfs2(G[v][i], v);
	}
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> n >> k;
	ll u, v;
	rep(i, 1, n-1){
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	rep(i, 1, n) cin >> a[i];
	
	dfs(1, 0), dfs2(1, 0);
	
	/*rep(i, 1, n) for(auto u : G[i]){
		cout << i << " " << u << " "; outa(get(i, u), 0, 2*k-1);
	}*/
	
	rep(i, 1, n){
		ll sum[45] = {};
		for(auto u : G[i]){
			rep(j, 0, 2*k-1) sum[(j+1)%(2*k)] ^= get(i, u)[j];
		}
		ll ans = 0;
		rep(j, k, 2*k-1) ans ^= sum[j];
		if(ans) cout << 1 << " ";
		else cout << 0 << " ";
	}
	cout << endl;
	
	return 0;
}