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>
using namespace std;
const int maxn = 300001;
int n;
struct info{
	int v, w;
};
vector < info > edge[maxn];
int color[maxn];
void dfs(int now, int dis){
	if(color[now])return;
	color[now] = (dis & 1) + 1;
	for(auto &v : edge[now]){
		dfs(v.v, dis + v.w);
	}
}
int main(){
	cin >> n;
	for(int i = 1, a, b, w;i < n;++i){
            cin >> a >> b >> w;
		edge[a].push_back({b, w});
		edge[b].push_back({a, w});
	}
	dfs(1, 0);
	for(int i = 1;i <= n;++i)cout << color[i] - 1 << '\n';
}