#include<bits/stdc++.h>
using namespace std;
vector<pair<int,bool>> g[101111];
bool vis[201111];
bool col[101111];
void dfs(int x)
{
    vis[x] = 1;
    for(int i=0;i<g[x].size();++i)
    {
        if(!vis[g[x][i].first])
        {
            col[g[x][i].first] = col[x] ^ g[x][i].second;
            dfs(g[x][i].first);
        }
    }
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<n;++i)
    {
        int x,y,w;
        cin>>x>>y>>w;
        g[x].push_back({y,w%2});
        g[y].push_back({x,w%2});
    }
    dfs(1);
    for(int i=1;i<=n;++i)
    {
        cout<<col[i]<<endl;
    }
}
### Prompt

Develop a solution in cpp 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<bits/stdc++.h>
using namespace std;
vector<pair<int,bool>> g[101111];
bool vis[201111];
bool col[101111];
void dfs(int x)
{
    vis[x] = 1;
    for(int i=0;i<g[x].size();++i)
    {
        if(!vis[g[x][i].first])
        {
            col[g[x][i].first] = col[x] ^ g[x][i].second;
            dfs(g[x][i].first);
        }
    }
}
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<n;++i)
    {
        int x,y,w;
        cin>>x>>y>>w;
        g[x].push_back({y,w%2});
        g[y].push_back({x,w%2});
    }
    dfs(1);
    for(int i=1;i<=n;++i)
    {
        cout<<col[i]<<endl;
    }
}
```