#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll pp[100010];
vector<pair<int, ll> > g[100010];

void dfs(int u, int p ){
	for( auto pv:g[u] ) if( pv.first != p ){
		pp[pv.first] = pp[u] + pv.second;
		dfs(pv.first, u);
	}
}

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

### 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
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll pp[100010];
vector<pair<int, ll> > g[100010];

void dfs(int u, int p ){
	for( auto pv:g[u] ) if( pv.first != p ){
		pp[pv.first] = pp[u] + pv.second;
		dfs(pv.first, u);
	}
}

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

```