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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n,p1,p2,p3,col[100001];
vector<pair<int,int> > arr[100001];
void dfs(int cur,int pre,int c) {
    col[cur]=c;
    for (int i=0; i<arr[cur].size(); i++) {
        if (arr[cur][i].first==pre) continue;
        if (arr[cur][i].second%2==0) dfs(arr[cur][i].first,cur,c);
        else dfs(arr[cur][i].first,cur,!c);
    }
}
int main() {
    cin>>n;
    for (int i=0; i<n-1; i++) {
        cin>>p1>>p2>>p3; p1--; p2--;
        arr[p1].push_back({p2,p3});
      	arr[p2].push_back({p1,p3});
    }
    dfs(0,-1,0);
    for (int i=0; i<n; i++) cout<<col[i]<<'\n';
}