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;
typedef pair<ll, ll> pii;

int n, m;
int x, y, z;
int ans = 0;
int a[100009];
vector<pii> g[100009];

void dfs(int node, int w) {
  w = w % 2;
  a[node] = w;
  for(auto i:g[node]) {
    if(a[i.first]==-1) {
      // cout << i.first << endl;
      dfs(i.first, w + i.second);
    }
  }
}

int main() {
  memset(a, -1, sizeof(a));
  cin >> n;
  for(int i=0; i<n-1; i++) {
    cin >> x >> y >> z;
    z = z % 2;
    g[x].push_back(pii(y, z));
    g[y].push_back(pii(x, z));
  }
  dfs(1, 0);
  for(int i=1; i<=n; i++) {
    cout << a[i] << endl;
  }
}