#include<cstdio>
#include<vector>
#include<utility>
using namespace std;
typedef long long ll;
int n,v,u;
ll w,dis[100001];
vector<pair<int,int> > g[100001];

void dfs(int x,int y){
    for(int i=0;i<g[x].size();i++){
        if(g[x][i].first!=y){
            dis[g[x][i].first]=g[x][i].second+dis[x];
            dfs(g[x][i].first,x);
        }
    }
}

int main(){
    scanf("%d",&n);
    for(int i=0;i<n-1;i++){
        scanf("%d %d %lld",&v,&u,&w);
        g[u].push_back(make_pair(v,w));
        g[v].push_back(make_pair(u,w));
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++){
        printf("%lld\n",dis[i]%2);
    }
}
### 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<cstdio>
#include<vector>
#include<utility>
using namespace std;
typedef long long ll;
int n,v,u;
ll w,dis[100001];
vector<pair<int,int> > g[100001];

void dfs(int x,int y){
    for(int i=0;i<g[x].size();i++){
        if(g[x][i].first!=y){
            dis[g[x][i].first]=g[x][i].second+dis[x];
            dfs(g[x][i].first,x);
        }
    }
}

int main(){
    scanf("%d",&n);
    for(int i=0;i<n-1;i++){
        scanf("%d %d %lld",&v,&u,&w);
        g[u].push_back(make_pair(v,w));
        g[v].push_back(make_pair(u,w));
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++){
        printf("%lld\n",dis[i]%2);
    }
}
```