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:
/*input
5
2 5 2
2 3 10
1 3 8
3 4 2
*/
#include "bits/stdc++.h"
using namespace std;
vector<pair<long long,long long>> v[100005];
int a[100005];
void dfs(int u, int dad, int c){
	a[u] = c;
	for(auto i:v[u]) if(i.first!=dad){
		dfs(i.first,u,c^(i.second&1));
	}
}
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0); cout.tie(0);
	int n;
	cin>>n;
	for(int i=n-1,x,y,p;i--;){
		cin>>x>>y>>p;
		v[x].push_back({y,p});
		v[y].push_back({x,p});
	}
	dfs(1,0,0);
	for(int i=1;i<=n;cout<<a[i++]<<'\n');
}