#include<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int n,col[N];

int cnt=1,head[N],nxt[N<<1],to[N<<1],len[N<<1];
void add_edge(int x,int y,int z) {
	nxt[++cnt]=head[x]; to[cnt]=y; len[cnt]=z; head[x]=cnt;
}

void dfs(int x) {
	for (int i=head[x];i;i=nxt[i]) {
		int y=to[i];
		if (!col[y]) {
			if (len[i]%2==0) col[y]=col[x]; else col[y]=3-col[x];
			dfs(y);
		}
	}
}

int main()
{
	cin>>n;
	for (int i=1;i<n;i++) {
		int x,y,z; scanf("%d%d%d",&x,&y,&z);
		add_edge(x,y,z); 
		add_edge(y,x,z);
	}
	col[1]=1;
	dfs(1);
	for (int i=1;i<=n;i++) printf("%d\n",col[i]-1);
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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<bits/stdc++.h>
using namespace std;
const int N=1e5+10;
int n,col[N];

int cnt=1,head[N],nxt[N<<1],to[N<<1],len[N<<1];
void add_edge(int x,int y,int z) {
	nxt[++cnt]=head[x]; to[cnt]=y; len[cnt]=z; head[x]=cnt;
}

void dfs(int x) {
	for (int i=head[x];i;i=nxt[i]) {
		int y=to[i];
		if (!col[y]) {
			if (len[i]%2==0) col[y]=col[x]; else col[y]=3-col[x];
			dfs(y);
		}
	}
}

int main()
{
	cin>>n;
	for (int i=1;i<n;i++) {
		int x,y,z; scanf("%d%d%d",&x,&y,&z);
		add_edge(x,y,z); 
		add_edge(y,x,z);
	}
	col[1]=1;
	dfs(1);
	for (int i=1;i<=n;i++) printf("%d\n",col[i]-1);
	return 0;
}
```