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 ine long long

using namespace std;

vector <pair<int,int> > v[100010];
bool vis[100010];
int c[100010];

void dfs(int n, int d)
{
	if(d%2)
		c[n] = 1;

	vis[n] = true;

	for(auto i:v[n])
	{
		if(!vis[i.first])
			dfs(i.first, d+i.second);
	}
}

int32_t main()
{
	int n;
	cin >> n;

	int x,y,z;

	for(int i=0;i<n;i++)
	{
		cin >> x >> y >> z;
		v[x].push_back({y,z});
		v[y].push_back({x,z});
	}

	dfs(1,0);

	for(int i=1;i<=n;i++)
		cout << c[i] << "\n";
}