#include<bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define pii pair<int,int>

int N;
vector<pii> G[100010];
int ans[100010];

void DFS(int v, int d, int p=0)
{
    ans[v] = d;
    for(auto e : G[v])if(e.first != p)
    {
        DFS(e.first, e.second ^ d, v);
    }
}

signed main()
{
    cin >> N;
    rep(i, N - 1)
    {
        int u, v, w;
        cin >> u >> v >> w;
        w = w % 2;
        G[u].push_back({v, w});
        G[v].push_back({u, w});
    }

    DFS(1, 0);

    for(int i = 1; i <= N; i++)cout << ans[i] << endl;

    return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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>
using namespace std;

#define int long long
#define rep(i,n) for(int i=0;i<(n);i++)
#define pii pair<int,int>

int N;
vector<pii> G[100010];
int ans[100010];

void DFS(int v, int d, int p=0)
{
    ans[v] = d;
    for(auto e : G[v])if(e.first != p)
    {
        DFS(e.first, e.second ^ d, v);
    }
}

signed main()
{
    cin >> N;
    rep(i, N - 1)
    {
        int u, v, w;
        cin >> u >> v >> w;
        w = w % 2;
        G[u].push_back({v, w});
        G[v].push_back({u, w});
    }

    DFS(1, 0);

    for(int i = 1; i <= N; i++)cout << ans[i] << endl;

    return 0;
}

```