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;
typedef long long ll;
const ll inf=(1ll<<61);
const ll mod=1e9+7;
const int MX=3e5+9;
int n,x,y,z;
vector<pair<int,int> >v[MX];
int ans[MX];
void dfs(int x,int p,int d){
    ans[x]=d%2;
    for(auto pp:v[x]){
        if(pp.first==p)continue;
        dfs(pp.first,x,pp.second+d);
    }
}
int main(){
    cin>>n;
    for(int i=0;i<n-1;i++){
        scanf("%d%d%d",&x,&y,&z);
        z%=2;
        v[x].push_back({y,z});
        v[y].push_back({x,z});
    }
    dfs(1,-1,0);
    for(int i=1;i<=n;i++){
        cout<<ans[i]<<endl;
    }
}