#include "bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
typedef long long ll;
vector<pair<int, int> > conn[101010];
int ans[101010];
void dfs(int a, int pa, int c)
{
	ans[a] = c;
	for(auto x: conn[a])
		if(x.first != pa)
			dfs(x.first, a, c^x.second);
}
int main()
{
    int N;
	cin>>N;
	rep(i,N-1){
		int u, v, w;
		cin>>u>>v>>w;
		conn[u].emplace_back(v, w%2);
		conn[v].emplace_back(u, w%2);
	}
	dfs(1, 0, 0);
	rep(i,N)
		cout<<ans[i+1]<<endl;
}
### Prompt

Please formulate 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 "bits/stdc++.h"
using namespace std;
#define rep(i,n) for(int i=0;i<(n);i++)
typedef long long ll;
vector<pair<int, int> > conn[101010];
int ans[101010];
void dfs(int a, int pa, int c)
{
	ans[a] = c;
	for(auto x: conn[a])
		if(x.first != pa)
			dfs(x.first, a, c^x.second);
}
int main()
{
    int N;
	cin>>N;
	rep(i,N-1){
		int u, v, w;
		cin>>u>>v>>w;
		conn[u].emplace_back(v, w%2);
		conn[v].emplace_back(u, w%2);
	}
	dfs(1, 0, 0);
	rep(i,N)
		cout<<ans[i+1]<<endl;
}
```