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 N=1e5+5,MOD=1e9+7;

typedef long long ll;

int n,c,u,v,w,k,tc,ans;

vector<pair<int,int> > g[N];

int color[N];

void dfs(int u,int p){
	for(auto pp : g[u]){

		int v = pp.first , cst = pp.second;
		if(v == p)
            continue;
		if(cst%2)
			color[v]=!color[u];
		else
			color[v]=color[u];
		dfs(v,u);
	}
}

int main() {
    scanf("%d",&n);
	for(int i=1;i<n;i++){
        scanf("%d%d%d",&u,&v,&w);
        g[--u].push_back({--v,w});
		g[v].push_back({u,w});
	}
	color[1]=0;
	dfs(0,0);

	for(int i=0;i<n;i++)
        printf("%d\n",color[i] );
}