#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define li long long
using namespace std;
int n,tot;
int hd[100009],eg[200009],nxt[200009],len[200009];
int cl[100009];
void ins(int a,int b,int c)
{
	eg[++tot]=b;
	len[tot]=c;
	nxt[tot]=hd[a];
	hd[a]=tot;
}
void dfs(int x,int fa,li dis)
{
	if(dis&1)
		cl[x]=1;
	else
		cl[x]=0;
	for(int i=hd[x];i;i=nxt[i])
		if(eg[i]!=fa)
			dfs(eg[i],x,dis+len[i]);
}
int main()
{
	scanf("%d",&n);
	for(int i=1,a,b,c;i<=n-1;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		ins(a,b,c);
		ins(b,a,c);
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++)
		printf("%d\n",cl[i]);
	return 0;
}
### Prompt

Your challenge is to write a CPP solution to the following 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

### Response

```cpp
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define li long long
using namespace std;
int n,tot;
int hd[100009],eg[200009],nxt[200009],len[200009];
int cl[100009];
void ins(int a,int b,int c)
{
	eg[++tot]=b;
	len[tot]=c;
	nxt[tot]=hd[a];
	hd[a]=tot;
}
void dfs(int x,int fa,li dis)
{
	if(dis&1)
		cl[x]=1;
	else
		cl[x]=0;
	for(int i=hd[x];i;i=nxt[i])
		if(eg[i]!=fa)
			dfs(eg[i],x,dis+len[i]);
}
int main()
{
	scanf("%d",&n);
	for(int i=1,a,b,c;i<=n-1;i++)
	{
		scanf("%d%d%d",&a,&b,&c);
		ins(a,b,c);
		ins(b,a,c);
	}
	dfs(1,0,0);
	for(int i=1;i<=n;i++)
		printf("%d\n",cl[i]);
	return 0;
}
```