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;
struct edge{int to,cost;};
vector<edge>G[100000];
int color[100000];
void dfs(int v,int c){
    color[v]=c;
    for(int i=0;i<G[v].size();i++){
        if(color[G[v][i].to]==-1)
            dfs(G[v][i].to,(c+G[v][i].cost)%2);
    }
}
int main(){
    int n,u,v,w;
    cin>>n;
    while(cin>>u>>v>>w){
        G[u-1].push_back((edge){v-1,w});
        G[v-1].push_back((edge){u-1,w});
    }
    memset(color,-1,sizeof(color));
    dfs(0,0);
    for(int i=0;i<n;i++)
        cout<<color[i]<<endl;
}