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 <stdio.h>
#include <algorithm>
#include <vector>
using namespace std;
using ll = long long;

using pii = pair<int, ll>;
int n;
ll d[100010];
vector<pii> e[100010];
void dfs(int x, ll p) {
	if (d[x])
		return;
	d[x] = p;
	for (pii next : e[x])
		dfs(next.first, p + next.second);
	return;
}
int main() {
	scanf("%d", &n);
	for (int i = 0; i < n - 1; i++) {
		int u, v, w;
		scanf("%d %d %d", &u, &v, &w);
		e[u].emplace_back(v, w);
		e[v].emplace_back(u, w);
	}
	dfs(1, 1);
	for (int i = 1; i <= n; i++)
		printf("%d\n", d[i] & 1);
}