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

typedef long long ll;
ll n, ans[N];
bool v[N];
vector<ll> gr[N], ga[N];

void f(ll p, ll q) {
	ll i;
	if (v[p]) return;
	v[p] = 1;
	if (q % 2 == 0) ans[p] = 1;
	for (i = 0; i < gr[p].size(); i++) {
		f(gr[p][i], q + ga[p][i]);
	}
}

int main()
{
	ll i, t1, t2, t3;
	cin >> n;
	for (i = 1; i < n; i++) {
		scanf("%lld %lld %lld", &t1, &t2, &t3);
		gr[t1].push_back(t2); ga[t1].push_back(t3);
        gr[t2].push_back(t1); ga[t2].push_back(t3);
	}
	f(1, 0);
	for (i = 1; i <= n; i++) printf("%d ", ans[i]);
    return 0;
}
### Prompt

Please formulate 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>
#define N 100005
using namespace std;

typedef long long ll;
ll n, ans[N];
bool v[N];
vector<ll> gr[N], ga[N];

void f(ll p, ll q) {
	ll i;
	if (v[p]) return;
	v[p] = 1;
	if (q % 2 == 0) ans[p] = 1;
	for (i = 0; i < gr[p].size(); i++) {
		f(gr[p][i], q + ga[p][i]);
	}
}

int main()
{
	ll i, t1, t2, t3;
	cin >> n;
	for (i = 1; i < n; i++) {
		scanf("%lld %lld %lld", &t1, &t2, &t3);
		gr[t1].push_back(t2); ga[t1].push_back(t3);
        gr[t2].push_back(t1); ga[t2].push_back(t3);
	}
	f(1, 0);
	for (i = 1; i <= n; i++) printf("%d ", ans[i]);
    return 0;
}
```