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;
typedef long long int ll;
typedef pair<ll,ll> ii;
typedef vector<ii> vii;
const int Nmaxx = 1e5 + 5;

vector<vii> g(Nmaxx);
vector<ll> ans(Nmaxx);

void dfs(int u, int p, bool color)
{
	for(ii b : g[u])
	{
		if(b.first == p )
			continue;
		ans[b.first] = (int)((!(b.second&1))?color:!color);
		dfs(b.first,u,ans[b.first]);
	}
}

int main()
{
	int n; cin>>n;
	for(int i = 1; i <n ; i++)
	{
		ll a,b,w; cin>>a>>b>>w;
		g[a].push_back({b,w});
		g[b].push_back({a,w});
	}
	dfs(1,0,false);
	for(int i = 1; i <=n ;i++)
		cout<<ans[i]<<endl;
	return 0;
}