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 int N = 3e5 + 10;
typedef pair<int, int> pa;
vector<pa> g[N];
int f[N];
void dfs(int fa, int u, int w) {
    f[u] = w;
    for(auto e : g[u]) {
        if(e.first == fa) continue;
        dfs(u, e.first, w^(e.second&1));
    }
}
int main() {
    int n;
    cin >> n;
    for(int i = 1, u, v, w; i < n; i++) {
        cin >> u >> v >> w;
        g[u].push_back(pa(v, w));
        g[v].push_back(pa(u, w));
    }
    dfs(-1, 1, 0);
    for(int i = 1; i <= n; i++) {
        cout << f[i] << '\n';
    }
    return 0;
}