#include <bits/stdc++.h>
using namespace std;
#define REP(i,n) for (int i=0;i<(n);++i)

int N,u,v,w;
vector<pair<int,int> > G[100010];
int color[100010];

// 奇数だったら今とは違う色で塗る。
void dfs(int v, int pv, int c){
    color[v] = c;
    for(auto nv : G[v]){
        if(nv.first == pv) continue;
        dfs(nv.first, v, c^nv.second);
    }
}

int main(){
    cin >> N;
    REP(i,N-1){
        cin >> u >> v >> w;
        G[u].push_back(make_pair(v,w%2));
        G[v].push_back(make_pair(u,w%2));
    }
    dfs(1,0,0);
    REP(i,N) cout << color[i+1] << "\n";
    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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 REP(i,n) for (int i=0;i<(n);++i)

int N,u,v,w;
vector<pair<int,int> > G[100010];
int color[100010];

// 奇数だったら今とは違う色で塗る。
void dfs(int v, int pv, int c){
    color[v] = c;
    for(auto nv : G[v]){
        if(nv.first == pv) continue;
        dfs(nv.first, v, c^nv.second);
    }
}

int main(){
    cin >> N;
    REP(i,N-1){
        cin >> u >> v >> w;
        G[u].push_back(make_pair(v,w%2));
        G[v].push_back(make_pair(u,w%2));
    }
    dfs(1,0,0);
    REP(i,N) cout << color[i+1] << "\n";
    return 0;
}
```