#include <iostream>
#include <cstdio>
#include <vector>
#define N 100005
using namespace std;

int n, ans[N], s[N];
vector<int> a[N], b[N];
void f(int p, int q) {
	int i, z;
	if (s[p] % 2 == 1) ans[p] = 1;
	for (i = 0; i < a[p].size(); i++) {
		z = a[p][i];
		if (z == q) continue;
		s[z] = s[p] + b[p][i];
		f(z, p);
	}
}

int main()
{
	int i, t1, t2, t3;
	cin >> n;
	for (i = 1; i <= n - 1; i++) {
		scanf ("%d %d %d", &t1, &t2, &t3);
		t3 %= 2;
		a[t1].push_back(t2);
		a[t2].push_back(t1);
		b[t1].push_back(t3);
		b[t2].push_back(t3);
	}
	f(1, 0);
	for (i = 1; i <= n; i++) printf("%d\n", ans[i]);
    return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
#define N 100005
using namespace std;

int n, ans[N], s[N];
vector<int> a[N], b[N];
void f(int p, int q) {
	int i, z;
	if (s[p] % 2 == 1) ans[p] = 1;
	for (i = 0; i < a[p].size(); i++) {
		z = a[p][i];
		if (z == q) continue;
		s[z] = s[p] + b[p][i];
		f(z, p);
	}
}

int main()
{
	int i, t1, t2, t3;
	cin >> n;
	for (i = 1; i <= n - 1; i++) {
		scanf ("%d %d %d", &t1, &t2, &t3);
		t3 %= 2;
		a[t1].push_back(t2);
		a[t2].push_back(t1);
		b[t1].push_back(t3);
		b[t2].push_back(t3);
	}
	f(1, 0);
	for (i = 1; i <= n; i++) printf("%d\n", ans[i]);
    return 0;
}

```