#include <bits/stdc++.h>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

vector<P>E[200000];
int d[200000];

void dfs(int v,int p){
	for(auto u:E[v]){
		if(u.second!=p){
			d[u.second]=(d[v]+u.first)%2;
			dfs(u.second,v);
		}
	}
}

int main(){
	int n;cin>>n;
	rep(i,n-1){
		int a,b,c;scanf("%d%d%d",&a,&b,&c);a--;b--;
		E[a].push_back(P(c,b));
		E[b].push_back(P(c,a));
	}
	dfs(0,-1);
	rep(i,n){
		printf("%d\n",d[i]);
	}
}
### Prompt

Generate 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>
#define rep(i,n)for(int i=0;i<(n);i++)
using namespace std;
typedef long long ll;
typedef pair<int,int>P;

vector<P>E[200000];
int d[200000];

void dfs(int v,int p){
	for(auto u:E[v]){
		if(u.second!=p){
			d[u.second]=(d[v]+u.first)%2;
			dfs(u.second,v);
		}
	}
}

int main(){
	int n;cin>>n;
	rep(i,n-1){
		int a,b,c;scanf("%d%d%d",&a,&b,&c);a--;b--;
		E[a].push_back(P(c,b));
		E[b].push_back(P(c,a));
	}
	dfs(0,-1);
	rep(i,n){
		printf("%d\n",d[i]);
	}
}
```