#include <cstdio>
#include <vector>

using namespace std;

const int V = 100010;

vector<pair<int,int>> graph[V];
bool visited[V];
int dist[V];

void dfs(int v,int d){
	if(visited[v])return;
	dist[v]=d%2;
	visited[v]=true;
	for(auto i:graph[v])dfs(i.first,(d+i.second)%2);
	//firstがpairの1要素目を指す
}

int main(void){
	int n;
	scanf("%d",&n);
	int u,v;
	long long w;
	for(int i=0;i<n-1;i++){
		scanf("%d%d%lld",&u,&v,&w);
		u--;
		v--;
		graph[u].push_back({v,w});//末尾に追加
		graph[v].push_back({u,w});
	}
	dfs(0,0);
	for(int i=0;i<n;i++){
		printf("%d\n",dist[i]%2);
	}
}

### Prompt

Please provide a cpp coded solution 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 <cstdio>
#include <vector>

using namespace std;

const int V = 100010;

vector<pair<int,int>> graph[V];
bool visited[V];
int dist[V];

void dfs(int v,int d){
	if(visited[v])return;
	dist[v]=d%2;
	visited[v]=true;
	for(auto i:graph[v])dfs(i.first,(d+i.second)%2);
	//firstがpairの1要素目を指す
}

int main(void){
	int n;
	scanf("%d",&n);
	int u,v;
	long long w;
	for(int i=0;i<n-1;i++){
		scanf("%d%d%lld",&u,&v,&w);
		u--;
		v--;
		graph[u].push_back({v,w});//末尾に追加
		graph[v].push_back({u,w});
	}
	dfs(0,0);
	for(int i=0;i<n;i++){
		printf("%d\n",dist[i]%2);
	}
}

```