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 SZ=100005;
vector<int> ev[SZ],od[SZ];
int ans[SZ];
void dfs(int x, int b, int c){
	ans[x]=c;
	for(int nxt:ev[x]){
		if(nxt==b) continue;
		dfs(nxt,x,c);
	}
	c=(c==1?0:1);
	for(int nxt:od[x]){
		if(nxt==b) continue;
		dfs(nxt,x,c);
	}
}
int main(void){
	int n; scanf("%d",&n);
	for(int i=1;i<n;i++){
		int a,b,c; scanf("%d %d %d",&a,&b,&c);
		if(c%2==0){
			ev[a].push_back(b);
			ev[b].push_back(a);
		}
		else{
			od[a].push_back(b);
			od[b].push_back(a);
		}
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++) printf("%d\n",ans[i]);
	return 0;
}