#include <bits/stdc++.h>
/// TONI BO$$ was here
/// #MLC

using namespace std;

struct much
{
    int x,v;
};

vector <much> G[100001];
int f[100001],d[100001];

void dfs(int i,int k)
{
    f[i]=1;
    d[i]=k;
    for(auto it : G[i])
        if(!f[it.x])
            dfs(it.x,k+it.v);
}

int main()
{
    int i,n,x,y,u;
    scanf("%d",&n);
    for(i=1; i<n; i++)
    {
        scanf("%d%d%d",&x,&y,&u);
        G[x].push_back({y,u%2});
        G[y].push_back({x,u%2});
    }
    dfs(1,0);
    for(i=1; i<=n; i++)
    printf("%d\n",d[i]%2);

    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>
/// TONI BO$$ was here
/// #MLC

using namespace std;

struct much
{
    int x,v;
};

vector <much> G[100001];
int f[100001],d[100001];

void dfs(int i,int k)
{
    f[i]=1;
    d[i]=k;
    for(auto it : G[i])
        if(!f[it.x])
            dfs(it.x,k+it.v);
}

int main()
{
    int i,n,x,y,u;
    scanf("%d",&n);
    for(i=1; i<n; i++)
    {
        scanf("%d%d%d",&x,&y,&u);
        G[x].push_back({y,u%2});
        G[y].push_back({x,u%2});
    }
    dfs(1,0);
    for(i=1; i<=n; i++)
    printf("%d\n",d[i]%2);

    return 0;
}

```