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

const int maxn = int(1e5)+5;

int vis[maxn], col[maxn];
vector<pair<int, int>> graph[maxn];

void dfs(int node, int d)
{
	if(vis[node]) return;
	col[node] = d%2;
	vis[node] = 1;
	for(auto it: graph[node]) dfs(it.first, (d+it.second)%2);
}

int main(void)
{
	int n, u, v, w;
	scanf("%d", &n);
	for(int i = 1;i < n;i++)
	{
		scanf("%d%d%d", &u, &v, &w);
		u--, v--;
		graph[u].push_back({v, w}), graph[v].push_back({u, w});
	}

	dfs(0, 0);
	for(int i = 0;i < n;i++) printf("%d\n", col[i]);
}
### Prompt

Your task is to create 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 <vector>
using namespace std;

const int maxn = int(1e5)+5;

int vis[maxn], col[maxn];
vector<pair<int, int>> graph[maxn];

void dfs(int node, int d)
{
	if(vis[node]) return;
	col[node] = d%2;
	vis[node] = 1;
	for(auto it: graph[node]) dfs(it.first, (d+it.second)%2);
}

int main(void)
{
	int n, u, v, w;
	scanf("%d", &n);
	for(int i = 1;i < n;i++)
	{
		scanf("%d%d%d", &u, &v, &w);
		u--, v--;
		graph[u].push_back({v, w}), graph[v].push_back({u, w});
	}

	dfs(0, 0);
	for(int i = 0;i < n;i++) printf("%d\n", col[i]);
}
```