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>
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
int n,a[100005];
vector<pii> g[100005];
void dfs(int x,int y){
	a[x]=y;
	for(int i=0;i<g[x].size();i++)
		if(a[g[x][i].F]==-1)dfs(g[x][i].F,(y+g[x][i].S%2)%2);
}
int main(){
	memset(a,-1,sizeof(a));
	cin >> n;
	for(int i=0;i<n-1;i++){
		int x,y,z;
		cin >> x >> y >> z;
		x--,y--;
		g[x].push_back(make_pair(y,z));
		g[y].push_back(make_pair(x,z));
	}
	for(int i=0;i<n;i++)
		if(a[i]==-1)dfs(i,0);
	for(int i=0;i<n;i++)
		cout << a[i] << endl;
    return 0;
}