#include <bits/stdc++.h>
#define rep(i,a) for(int i=0; i<int(a); i++)
using namespace std;

const int N=100005;
int n;
vector<pair<int,bool>> adj[N];
int ans[N];

void dfs(int u, int e) {
    for(auto p:adj[u]) {
        int v=p.first;
        if(v==e) continue;
        ans[v]=(p.second? (!ans[u]):ans[u]);
        dfs(v,u);
    }
}

int main() {
    cin >> n;
    rep(i,n-1) {
        int a,b,w;
        cin>>a>>b>>w;
        a--;b--;
        adj[a].push_back({b,w%2});
        adj[b].push_back({a,w%2});
    }
    ans[0]=0;
    dfs(0,-1);
    rep(i,n) cout << ans[i] << endl;
    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
#define rep(i,a) for(int i=0; i<int(a); i++)
using namespace std;

const int N=100005;
int n;
vector<pair<int,bool>> adj[N];
int ans[N];

void dfs(int u, int e) {
    for(auto p:adj[u]) {
        int v=p.first;
        if(v==e) continue;
        ans[v]=(p.second? (!ans[u]):ans[u]);
        dfs(v,u);
    }
}

int main() {
    cin >> n;
    rep(i,n-1) {
        int a,b,w;
        cin>>a>>b>>w;
        a--;b--;
        adj[a].push_back({b,w%2});
        adj[b].push_back({a,w%2});
    }
    ans[0]=0;
    dfs(0,-1);
    rep(i,n) cout << ans[i] << endl;
    return 0;
}
```