#include<cstdio>
#include<vector>
using namespace std;

#define i64 long long int
#define ran 101101

int n;
vector<pair<int, int> > e[ran];
int ans[ran];

void dfs(int cur, int last, i64 d) {
	ans[cur] = d%2;
	for(int i=0;i<(int)e[cur].size();i++) {
		int nxt = e[cur][i].first, v = e[cur][i].second;
		if(nxt==last) continue;
		
		dfs(nxt, cur, d+v);
	}
}

int main() {
	scanf("%d", &n);
	for(int i=1;i<n;i++) {
		int x, y, v; scanf("%d%d%d", &x, &y, &v);
		e[x].push_back(make_pair(y, v));
		e[y].push_back(make_pair(x, v));
	}
	
	dfs(1, -1, 0);
	
	for(int i=1;i<=n;i++) printf("%d\n", ans[i]);
	
	return 0;
}
### Prompt

In CPP, your task is to solve 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<cstdio>
#include<vector>
using namespace std;

#define i64 long long int
#define ran 101101

int n;
vector<pair<int, int> > e[ran];
int ans[ran];

void dfs(int cur, int last, i64 d) {
	ans[cur] = d%2;
	for(int i=0;i<(int)e[cur].size();i++) {
		int nxt = e[cur][i].first, v = e[cur][i].second;
		if(nxt==last) continue;
		
		dfs(nxt, cur, d+v);
	}
}

int main() {
	scanf("%d", &n);
	for(int i=1;i<n;i++) {
		int x, y, v; scanf("%d%d%d", &x, &y, &v);
		e[x].push_back(make_pair(y, v));
		e[y].push_back(make_pair(x, v));
	}
	
	dfs(1, -1, 0);
	
	for(int i=1;i<=n;i++) printf("%d\n", ans[i]);
	
	return 0;
}
```