#include<bits/stdc++.h>
#define int long long
using namespace std;
const int maxn = 1e5 + 5;
int n;
vector<pair<int, int>> g[maxn];
bool mark[maxn];
int ans[maxn];
void dfs(int x, int d) {
	ans[x] = d % 2;
	mark[x] = true;
	for(auto i : g[x]) {
		int u = i.first;
		int s = i.second;
		if(!mark[u]) 
			dfs(u, d + s);
	}
}

int32_t main() {
	cin >> n;
	for(int i = 0; i < n - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		u--, v--;
		g[u].push_back({v, w});
		g[v].push_back({u, w});
	}
	dfs(0, 0);
	for(int i = 0; i < n; i++)
		cout << ans[i] << endl;
}
### Prompt

Please create a solution in CPP 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<bits/stdc++.h>
#define int long long
using namespace std;
const int maxn = 1e5 + 5;
int n;
vector<pair<int, int>> g[maxn];
bool mark[maxn];
int ans[maxn];
void dfs(int x, int d) {
	ans[x] = d % 2;
	mark[x] = true;
	for(auto i : g[x]) {
		int u = i.first;
		int s = i.second;
		if(!mark[u]) 
			dfs(u, d + s);
	}
}

int32_t main() {
	cin >> n;
	for(int i = 0; i < n - 1; i++) {
		int u, v, w;
		cin >> u >> v >> w;
		u--, v--;
		g[u].push_back({v, w});
		g[v].push_back({u, w});
	}
	dfs(0, 0);
	for(int i = 0; i < n; i++)
		cout << ans[i] << endl;
}
```