Problem: We have a tree with N vertices numbered 1 to N. The i-th edge in the tree connects Vertex u_i and Vertex v_i, and its length is w_i. Your objective is to paint each vertex in the tree white or black (it is fine to paint all vertices the same color) so that the following condition is satisfied:

* For any two vertices painted in the same color, the distance between them is an even number.



Find a coloring of the vertices that satisfies the condition and print it. It can be proved that at least one such coloring exists under the constraints of this problem.

Constraints

* All values in input are integers.
* 1 \leq N \leq 10^5
* 1 \leq u_i < v_i \leq N
* 1 \leq w_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N
u_1 v_1 w_1
u_2 v_2 w_2
.
.
.
u_{N - 1} v_{N - 1} w_{N - 1}


Output

Print a coloring of the vertices that satisfies the condition, in N lines. The i-th line should contain `0` if Vertex i is painted white and `1` if it is painted black.

If there are multiple colorings that satisfy the condition, any of them will be accepted.

Examples

Input

3
1 2 2
2 3 1


Output

0
0
1


Input

5
2 5 2
2 3 10
1 3 8
3 4 2


Output

1
0
1
0
1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define X first
#define Y second

using namespace std;

const int MAXN = 1e5 + 1;

vector<pair<int,int>> nxt[MAXN];
bool vis[MAXN];
bool col[MAXN];

void paint(int v) {
	vis[v] = 1;
	for(auto e : nxt[v]) {
		if(vis[e.X]) continue;
		col[e.X] = col[v] ^ (e.Y & 1);
		paint(e.X);
	}
}


int main() {
	ios_base::sync_with_stdio(0);
	
	int n;
	cin >> n;
	for(int i = 0; i < n; i++) {
		int x, y, w;
		cin >> x >> y >> w;
		nxt[x].push_back({y,w});
		nxt[y].push_back({x,w});
	}
	
	paint(1);
	
	for(int v = 1; v <= n; v++) {
		cout << col[v] << "\n";
	}
	
	return 0;
}