#include <bits/stdc++.h>

using namespace std;
vector<pair<int, int> > adj[100005];
int cost[100005];

void dfs(int u, int cur){
    if(cost[u] != -1) return;;
    cost[u] = cur;
    for(auto p:adj[u])
        dfs(p.first, (cur + p.second) % 2);
}
int main() {
    int n;
    scanf("%d", &n);
    for(int i=0; i<n-1; i++){
        int x, y, z;
        scanf("%d %d %d", &x, &y, &z);
        adj[x].push_back({y, z});
        adj[y].push_back({x, z});
    }
    memset(cost, -1, sizeof(cost));
    dfs(1, 0);
    for(int i=1; i<=n; i++){
        printf("%d\n", cost[i]);
    }

}
### Prompt

Create a solution in cpp for 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;
vector<pair<int, int> > adj[100005];
int cost[100005];

void dfs(int u, int cur){
    if(cost[u] != -1) return;;
    cost[u] = cur;
    for(auto p:adj[u])
        dfs(p.first, (cur + p.second) % 2);
}
int main() {
    int n;
    scanf("%d", &n);
    for(int i=0; i<n-1; i++){
        int x, y, z;
        scanf("%d %d %d", &x, &y, &z);
        adj[x].push_back({y, z});
        adj[y].push_back({x, z});
    }
    memset(cost, -1, sizeof(cost));
    dfs(1, 0);
    for(int i=1; i<=n; i++){
        printf("%d\n", cost[i]);
    }

}
```