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<iostream>
#include<algorithm>
#include<vector>
#include<math.h>
#include<iomanip>
using namespace std;
typedef long long ll;
#define N 100010
struct edge{
  ll to,cost;  
  edge(ll to,ll cost){
      this->to=to,this->cost=cost;
  }
};
ll n;
vector<edge> g[N];
ll d[N];
void dfs(ll x,ll from,ll dis){
    d[x]=dis;
    for(auto y:g[x])if(y.to!=from){
	dfs(y.to,x,dis+y.cost);
    }
}
int main(){
  cin>>n;
  for(int i=0;i<n-1;i++){
      ll a,b,c;cin>>a>>b>>c;
      g[a].push_back(edge(b,c));
      g[b].push_back(edge(a,c));
  }
  dfs(1,0,0);
  for(int i=1;i<=n;i++)cout<<d[i]%2<<endl;
  return 0;
}