Snuke Festival 2017 will be held in a tree with N vertices numbered 1,2, ...,N. The i-th edge connects Vertex a_i and b_i, and has joyfulness c_i.

The staff is Snuke and N-1 black cats. Snuke will set up the headquarters in some vertex, and from there he will deploy a cat to each of the other N-1 vertices.

For each vertex, calculate the niceness when the headquarters are set up in that vertex. The niceness when the headquarters are set up in Vertex i is calculated as follows:

* Let X=0.
* For each integer j between 1 and N (inclusive) except i, do the following:
* Add c to X, where c is the smallest joyfulness of an edge on the path from Vertex i to Vertex j.
* The niceness is the final value of X.

Constraints

* 1 \leq N \leq 10^{5}
* 1 \leq a_i,b_i \leq N
* 1 \leq c_i \leq 10^{9}
* The given graph is a tree.
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N
a_1 b_1 c_1
:
a_{N-1} b_{N-1} c_{N-1}


Output

Print N lines. The i-th line must contain the niceness when the headquarters are set up in Vertex i.

Examples

Input

3
1 2 10
2 3 20


Output

20
30
30


Input

15
6 3 2
13 3 1
1 13 2
7 1 2
8 1 1
2 8 2
2 12 2
5 2 2
2 11 2
10 2 2
10 9 1
9 14 2
4 14 1
11 15 2


Output

16
20
15
14
20
15
16
20
15
20
20
20
16
15
20


Input

19
19 14 48
11 19 23
17 14 30
7 11 15
2 19 15
2 18 21
19 10 43
12 11 25
3 11 4
5 19 50
4 11 19
9 12 29
14 13 3
14 6 12
14 15 14
5 1 6
8 18 13
7 16 14


Output

103
237
71
263
370
193
231
207
299
358
295
299
54
368
220
220
319
237
370

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long ll;
int N,A[100010],B[100010],E[100010] = {0},visited[200010] = {0};
ll C[100010],ans[200010] = {0};

vector<pair<ll,int>> e;
vector<vector<pair<ll,int>>> v(200010);

int p[100010] = {0},r[100010] = {0};
ll s[100010] = {0};

void init(int N){
	for(int i=0;i<=N;++i){
		p[i] = i;
		s[i] = 1;
		E[i] = i;
	}
}

int root(int a){
	if(p[a] == a) return a;
	return (p[a] = root(p[a]));
}

bool is_same_set(int a,int b){
	return root(a) == root(b);
}

void unite(int a, int b){
	a = root(a);
	b = root(b);
	if(a==b) return;
	if(r[a]<r[b]){
		p[a] = b;
		s[b] += s[a];
		s[a] = s[b];
	}else{
		p[b] = a;
		s[a] += s[b];
		s[b] = s[a];
		if(r[a] == r[b]) r[a]++;
	}
}

void dfs(int n){
	visited[n] = 1;
	for(int i=0;i<v[n].size();i++){
		int t = v[n][i].second;
		ll cost = v[n][i].first;
		if(visited[t]==0){
			ans[t] = ans[n] + cost;
			dfs(t);
		}
	}
}

int main(){
	cin >> N;
	init(N);
	if(N==1){
		cout << 0 << endl;
		return 0;
	}
	for(int i=1;i<=N-1;i++){
		cin >> A[i] >> B[i] >> C[i];
		e.push_back({C[i],i});
	}
	sort(e.begin(),e.end());
	for(int i=N-2;i>=0;i--){
		int t = e[i].second;
		ll cost = e[i].first;
		v[t+N].push_back({cost*s[root(B[t])],E[root(A[t])]});
		v[t+N].push_back({cost*s[root(A[t])],E[root(B[t])]});
		unite(A[t],B[t]);
		E[root(A[t])] = t+N;
	}
	dfs(N+e[0].second);
	for(int i=1;i<=N;i++){
		cout << ans[i] << endl;
	}
}