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

int ans[100005];
vector<pair<long long,long long>> tree[100005];
void dfs(int f,int now, long long cost) {
	ans[now] = cost % 2;
	for (int i = 0; i < tree[now].size(); i++) {
		if (tree[now][i].first== f) continue;
		dfs(now, tree[now][i].first, cost + tree[now][i].second);
	}
}
int main() {
	int n; cin >> n;
	for (int i = 0; i < n - 1; i++) {
		long long u, v, w; cin >> u >> v >> w;
		u--; v--;
		tree[u].push_back({ v,w });
		tree[v].push_back({ u,w });
	}
	dfs(-1, 0, 0);
	for (int i = 0; i < n; i++) {
		cout << ans[i] << endl;
	}
}
### 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<vector>
using namespace std;

int ans[100005];
vector<pair<long long,long long>> tree[100005];
void dfs(int f,int now, long long cost) {
	ans[now] = cost % 2;
	for (int i = 0; i < tree[now].size(); i++) {
		if (tree[now][i].first== f) continue;
		dfs(now, tree[now][i].first, cost + tree[now][i].second);
	}
}
int main() {
	int n; cin >> n;
	for (int i = 0; i < n - 1; i++) {
		long long u, v, w; cin >> u >> v >> w;
		u--; v--;
		tree[u].push_back({ v,w });
		tree[v].push_back({ u,w });
	}
	dfs(-1, 0, 0);
	for (int i = 0; i < n; i++) {
		cout << ans[i] << endl;
	}
}
```