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

ll n, vt[N], ans[N];
vector<ll> tr[N], tw[N];

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

int main() {
	ll i, u, v, w;
	cin >> n;
	for(i=0; i<n-1; i++) {
		scanf("%lld%lld%lld", &u, &v, &w);
		tr[u].push_back(v); tw[u].push_back(w);
		tr[v].push_back(u); tw[v].push_back(w);
	}
	f(1, 0);
	for(i=1; i<=n; i++) printf("%lld\n", ans[i]);
	return 0;
}

### Prompt

Your challenge is to write 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, vt[N], ans[N];
vector<ll> tr[N], tw[N];

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

int main() {
	ll i, u, v, w;
	cin >> n;
	for(i=0; i<n-1; i++) {
		scanf("%lld%lld%lld", &u, &v, &w);
		tr[u].push_back(v); tw[u].push_back(w);
		tr[v].push_back(u); tw[v].push_back(w);
	}
	f(1, 0);
	for(i=1; i<=n; i++) printf("%lld\n", ans[i]);
	return 0;
}

```