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 <iostream>
#include <cstdio>
#define maxn 100005
using namespace std;
int cnt,head[maxn],flag[maxn];
struct fdfdfd{int next,to,w;}e[maxn<<1];
void addedge(int u,int v,int w){e[++cnt].to=v; e[cnt].w=w; e[cnt].next=head[u]; head[u]=cnt;}
void dfs(int u,int pre,int op)
{
	flag[u]=op;
	for(int i=head[u];i;i=e[i].next)
	{
		int v=e[i].to,w=e[i].w;
		if(v==pre) continue;
		dfs(v,u,w&1?op^1:op);
	}
}
int main()
{
	int n; scanf("%d",&n);
	for(int i=1,u,v,w;i<n;++i) scanf("%d%d%d",&u,&v,&w),addedge(u,v,w),addedge(v,u,w);
	dfs(1,0,0);
	for(int i=1;i<=n;++i) printf("%d\n",flag[i]);
}