#include <bits/stdc++.h>
using namespace std;

#define MAX_N 100005

int n;
int a ,b ,c;
vector <pair<int ,int>> adj[MAX_N];

bool ans[MAX_N];
void dfs(int u ,int p=-1 ,int d=0){
    ans[u] = d&1;
    for(auto v : adj[u]) if(v.first != p)
        dfs(v.first ,u ,d+v.second);
}

int main()
{
    scanf("%d",&n);
    for(int i=1; i<n; i++){
        scanf("%d%d%d",&a,&b,&c);
        adj[a].push_back({b,c});
        adj[b].push_back({a,c});
    }

    dfs(1);
    for(int i=1; i<=n; i++)
        printf("%d\n",ans[i]);
}

### Prompt

Generate 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 <bits/stdc++.h>
using namespace std;

#define MAX_N 100005

int n;
int a ,b ,c;
vector <pair<int ,int>> adj[MAX_N];

bool ans[MAX_N];
void dfs(int u ,int p=-1 ,int d=0){
    ans[u] = d&1;
    for(auto v : adj[u]) if(v.first != p)
        dfs(v.first ,u ,d+v.second);
}

int main()
{
    scanf("%d",&n);
    for(int i=1; i<n; i++){
        scanf("%d%d%d",&a,&b,&c);
        adj[a].push_back({b,c});
        adj[b].push_back({a,c});
    }

    dfs(1);
    for(int i=1; i<=n; i++)
        printf("%d\n",ans[i]);
}

```