#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
typedef long long ll;
vector<pair<int,int> > v[100005];
bool vis[100005];
int col[100005];
void dfs(int x,int c){
    col[x] = c;
    vis[x] = 1;
    for(pair<int,int> p:v[x]){
        int y = p.fi;
        int w = p.se;
        if(vis[y])continue;
        if(w%2)dfs(y,1-c);
        else dfs(y,c);
    }
}
int main(){
    int n;
    cin>>n;
    for(int i=1;i<n;i++){
        int x,y,w;
        cin>>x>>y>>w;
        v[x].pb({y,w});
        v[y].pb({x,w});
    }
    dfs(1,0);
    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;
#define pb push_back
#define fi first
#define se second
typedef long long ll;
vector<pair<int,int> > v[100005];
bool vis[100005];
int col[100005];
void dfs(int x,int c){
    col[x] = c;
    vis[x] = 1;
    for(pair<int,int> p:v[x]){
        int y = p.fi;
        int w = p.se;
        if(vis[y])continue;
        if(w%2)dfs(y,1-c);
        else dfs(y,c);
    }
}
int main(){
    int n;
    cin>>n;
    for(int i=1;i<n;i++){
        int x,y,w;
        cin>>x>>y>>w;
        v[x].pb({y,w});
        v[y].pb({x,w});
    }
    dfs(1,0);
    for(int i=1;i<=n;i++)cout<<col[i]<<endl;
}

```