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=2e5+7;
int n,cnt,hd[N],v[N],nxt[N],w[N];
long long d[N];
void adde(int x,int y,int z){v[++cnt]=y,nxt[cnt]=hd[x],w[cnt]=z,hd[x]=cnt;}
void dfs(int u,int fa)
{
	for(int i=hd[u];i;i=nxt[i])
	if(v[i]!=fa)
	{
		d[v[i]]=d[u]+w[i];
		dfs(v[i],u);
	}
}
int main()
{
	scanf("%d",&n);
	for(int i=1,x,y,z;i<n;i++)scanf("%d%d%d",&x,&y,&z),adde(x,y,z),adde(y,x,z);
	dfs(1,0);
	for(int i=1;i<=n;i++)
	{
		int op=d[i]&1;
		printf("%d\n",op);
	}
}