#include<iostream>
#include<vector>
using namespace std;

typedef struct{int to,cost;}edge;
vector<edge>tree[100000];

long long depth[100000];

void dfs(int current_v,int parent,long long cost){
    depth[current_v]=cost;
    for(auto x:tree[current_v])if(x.to!=parent)dfs(x.to,current_v,cost+x.cost);//(木の子孫は全て未訪問)
    return;
}

int main(void){
    int n,u,v,w;
    cin>>n;
    for(int i=1;i<n;i++){
        cin>>u>>v>>w;
        tree[u-1].push_back(edge{v-1,w});
        tree[v-1].push_back(edge{u-1,w});
    }
    dfs(0,-1,0LL);
    for(int i=0;i<n;i++)cout<<depth[i]%2<<endl;
    return 0;
}
### 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<iostream>
#include<vector>
using namespace std;

typedef struct{int to,cost;}edge;
vector<edge>tree[100000];

long long depth[100000];

void dfs(int current_v,int parent,long long cost){
    depth[current_v]=cost;
    for(auto x:tree[current_v])if(x.to!=parent)dfs(x.to,current_v,cost+x.cost);//(木の子孫は全て未訪問)
    return;
}

int main(void){
    int n,u,v,w;
    cin>>n;
    for(int i=1;i<n;i++){
        cin>>u>>v>>w;
        tree[u-1].push_back(edge{v-1,w});
        tree[v-1].push_back(edge{u-1,w});
    }
    dfs(0,-1,0LL);
    for(int i=0;i<n;i++)cout<<depth[i]%2<<endl;
    return 0;
}
```