#import<bits/stdc++.h>
using namespace std;
typedef struct{int x;int y;int c;}M;
M g[200000];
int a,b,i,j,l,n,w,c[100005],p[100005],v[100005];
queue<int>z;
int Q(M a,M b){return a.x<b.x;}
int main()
{
for(scanf("%d",&n);l<n-1;l++)
{
scanf("%d%d%d",&g[l].x,&g[l].y,&g[l].c);g[l+n-1].x=g[l].y,g[l+n-1].y=g[l].x,g[l+n-1].c=g[l].c;
}
sort(g,g+n+n-2,Q);
for(;g[++i].x;)if(g[i].x>g[i-1].x)c[g[i].x]=i;c[n+1]=2*n-2;
for(z.push(1);z.size();z.pop())for(v[a=z.front()]=1,j=c[a];j<c[a+1];j++)
if(v[g[j].y]<1)z.push(g[j].y),p[g[j].y]=g[j].c%2?!p[a]:p[a];
for(i=0;i++<n;)printf("%d\n",p[i]);
}
### Prompt

Your challenge is to write 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
#import<bits/stdc++.h>
using namespace std;
typedef struct{int x;int y;int c;}M;
M g[200000];
int a,b,i,j,l,n,w,c[100005],p[100005],v[100005];
queue<int>z;
int Q(M a,M b){return a.x<b.x;}
int main()
{
for(scanf("%d",&n);l<n-1;l++)
{
scanf("%d%d%d",&g[l].x,&g[l].y,&g[l].c);g[l+n-1].x=g[l].y,g[l+n-1].y=g[l].x,g[l+n-1].c=g[l].c;
}
sort(g,g+n+n-2,Q);
for(;g[++i].x;)if(g[i].x>g[i-1].x)c[g[i].x]=i;c[n+1]=2*n-2;
for(z.push(1);z.size();z.pop())for(v[a=z.front()]=1,j=c[a];j<c[a+1];j++)
if(v[g[j].y]<1)z.push(g[j].y),p[g[j].y]=g[j].c%2?!p[a]:p[a];
for(i=0;i++<n;)printf("%d\n",p[i]);
}
```