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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef string str;

#define pb push_back
#define ppp pop_back
#define pii pair<int,int>
#define fi first
#define se second
#define stie std::tie
#define vec vector
#define	forn(i, l, r) for (int i=l; i<=r; i++)
#define forb(i, r, l) for (int i=r; i>=l; i--)
#define emp empty
#define beg begin
#define ins insert
#define cle clear
#define era erase
#define que queue
#define pque priority_queue
#define mset multiset
#define deq deque
#define sta stack
#define con const
#define rsz resize
#define ass assign
#define lowb lower_bound
#define uppb upper_bound

template<class T> void mini(T& a, T b) { a = min(a, b); }
template<class T> void maxi(T& a, T b) { a = max(a, b); }
template<class T1, class T2> ostream& operator<<(ostream &out, pair<T1, T2> a) { return out << a.first << " " << a.second; }
template<class T> ostream& operator<<(ostream& out, vector<T> a) { for (auto& i : a) out << i << " "; return out; }
template<class T> ostream& operator<<(ostream& out, set<T> a) { for (auto& i : a) out << i << " "; return out; }
template<class T> ostream& operator<<(ostream& out, multiset<T> a) { for (auto& i : a) out << i << " "; return out; }
template<class T1, class T2> ostream& operator<<(ostream& out, map<T1, T2> a) { for (auto& i : a) out << i << "\n"; return out; }
template<class T1, class T2> ostream& operator<<(ostream& out, unordered_map<T1, T2> a) { for (auto& i : a) out << i << "\n"; return out; }
template<class T> ostream& operator<<(ostream& out, queue<T> a) { while (!a.empty()) { out << a.front() << " "; a.pop(); } return out; }
template<class T> ostream& operator<<(ostream& out, deque<T> a) { while (!a.empty()) { out << a.front() << " "; a.pop_front(); } return out; }
template<class T> ostream& operator<<(ostream& out, priority_queue<T> a) { while (!a.empty()) { out << a.top() << "\n"; a.pop(); } return out; }
template<class T> void out(T a) { cout << a << endl; }
template<class T1, class T2> void out(T1 a, T2 b) { cout << a << " " << b << endl; }
template<class T1, class T2, class T3> void out(T1 a, T2 b, T3 c) { cout << a << " " << b << " " << c << endl; }
template<class T1, class T2, class T3, class T4> void out(T1 a, T2 b, T3 c, T4 d) { cout << a << " " << b << " " << c << " " << d << endl; }
template<class T> void out(vector<vector<T>> a) { for (vector<T> i : a) out(i); }
template<class T> void out_(T a[], int l, int r) { for (int i = l; i <= r; i++) cout << a[i] << " "; cout << endl; }
void out() { cout << "OK" << endl; }
template<class T> void sort(vector<T>& a) { sort(a.begin(), a.end()); }
template<class T> void uniq(vector<T>& a) { sort(a); a.erase(unique(a.begin(), a.end()), a.end()); }
template<class T> vector<T> set_int(vector<T> a, vector<T> b) { sort(a); sort(b); vector<T> res; set_intersection(a.begin(), a.end(), b.begin(), b.end(), back_inserter(res)); return res; }
clock_t start_time; void start_timer() { start_time = clock(); }
double get_time() { return (double)(clock() - start_time) / CLOCKS_PER_SEC; }
#if __SIZEOF_INT128__ >= 16
typedef __int128 LL;
istream& operator>>(istream& in, __int128& a) { int64_t b; cin >> b; a = b; return in; }
ostream& operator<<(ostream& out, const __int128 a) {
	unsigned __int128 b = a < 0 ? -a : a; char buf[128]; char* c = end(buf); do { --c; *c = "0123456789"[b % 10]; b /= 10; } while (b);
	if (a < 0) { --c; *c = '-'; } int len = end(buf) - c; out.rdbuf()->sputn(c, len); return out;
}
#endif

con int N = 1e5 + 5;
con int K = 25;

int a[N];
vec<int> g[N];
int dp[N][2 * K];
bool ans[N];
int k;

void dfs(int v = 1, int p = 0) {
	dp[v][0] = a[v];
	for (int u : g[v]) {
		if (u == p) continue;
		dfs(u, v);
		forn(i, 1, k-1) {
			dp[v][i] ^= dp[u][i-1];
		}
		dp[v][0] ^= dp[u][k-1];
	}
}

void dfs2(int v = 1, int p = 0, vec<int> cur = vec<int>(2 * K, 0)) {
	forn(i, 0, k-1) {
		cur[i] ^= dp[v][i];
	}
	int odd = 0;
	forn(i, k/2, k-1) {
		odd ^= cur[i];
	}
	ans[v] = odd;
	for (int u : g[v]) {
		if (u == p) continue;
		vec<int> nxt = cur;
		forn(i, 1, k-1) {
			nxt[i] ^= dp[u][i-1];
		}
		nxt[0] ^= dp[u][k-1];
		int last = nxt[k-1];
		forb(i, k-1, 1) {
			nxt[i] = nxt[i-1];
		}
		nxt[0] = last;
		dfs2(u, v, nxt);
	}
}

void Solve() {
	int n;
	cin >> n >> k;
	k *= 2;
	forn(i, 1, n-1) {
		int u, v;
		cin >> u >> v;
		g[u].pb(v);
		g[v].pb(u);
	}
	forn(i, 1, n) cin >> a[i];
	dfs();
	dfs2();
	forn(i, 1, n) {
		cout << ans[i] << " ";
	}
	cout << "\n";
}

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0);
#ifdef LOCAL
	freopen("input.txt", "r", stdin);
#endif
	start_timer();
	Solve();
	return 0;
}