#include <bits/stdc++.h>

using namespace std;

int n;
vector<pair<int,int> > G[100010];
long long dis[100010];
int ans[100010];

void f(int u, int p) {
    if(dis[u] % 2 == 0) ans[u] = 1;
    for (auto v: G[u]) {
        if(v.first == p) continue;
        dis[v.first] = dis[u] + v.second;
        f(v.first, u);
    }
}

int main() {
    
    cin >> n;

    for (int i = 0; i < n - 1; ++i) {
        int x,y,w;
        cin >> x >> y >> w;
        G[x].push_back({y,w});
        G[y].push_back({x,w});
    }
 
    f(1,1);

    for (int i = 1; i <= n; ++i)
        cout << ans[i] << "\n";

    return 0;
}

### Prompt

Generate a Cpp solution to the following 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

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;

int n;
vector<pair<int,int> > G[100010];
long long dis[100010];
int ans[100010];

void f(int u, int p) {
    if(dis[u] % 2 == 0) ans[u] = 1;
    for (auto v: G[u]) {
        if(v.first == p) continue;
        dis[v.first] = dis[u] + v.second;
        f(v.first, u);
    }
}

int main() {
    
    cin >> n;

    for (int i = 0; i < n - 1; ++i) {
        int x,y,w;
        cin >> x >> y >> w;
        G[x].push_back({y,w});
        G[y].push_back({x,w});
    }
 
    f(1,1);

    for (int i = 1; i <= n; ++i)
        cout << ans[i] << "\n";

    return 0;
}

```