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 <functional>
#include <algorithm>
#include <iostream>
#include <cassert>
#include <string>
#include <vector>
#include <cmath>
#include <tuple>
#include <queue>
#include <map>
using namespace std;
// ----------------------------------------------- Union FInd Query -------------------------------------

int group[1000009], num[1000009], size_, cnts; vector<int>GR[1000009];
void init() {
	for (int i = 0; i < size_; i++) { group[i] = i; num[i] = i; GR[i].push_back(i); }
	cnts = size_;
}
void unite(int p1, int p2) {
	int z1 = group[p1], z2 = group[p2]; if (z1 == z2)return;
	if (GR[z1].size() < GR[z2].size())swap(z1, z2);
	for (int i = 0; i < GR[z2].size(); i++) {
		GR[z1].push_back(GR[z2][i]); group[GR[z2][i]] = z1;
	}
	GR[z2].clear(); num[z1] = cnts; cnts++;
}
int Size(int p1) {
	return GR[group[p1]].size();
}
bool same(int p1, int p2) {
	if (group[p1] == group[p2])return true;
	return false;
}

// ----------------------------------------------- Main ------------------------------------------------

long long n, a[100009], b[100009], w[100009], p[300009];
vector<tuple<int, int, int>>V; vector<int>G, x[300009];

int main() {
	cin >> n;
	for (int i = 1; i <= n - 1; i++) {
		cin >> a[i] >> b[i] >> w[i]; V.push_back(make_tuple(w[i], a[i], b[i]));
	}
	sort(V.begin(), V.end(), greater<tuple<int, int, int>>());
	size_ = n + 1; init();
	for (int i = 0; i < V.size(); i++) {
		long long a1 = get<1>(V[i]), a2 = get<2>(V[i]), a3 = get<0>(V[i]);
		long long C1 = Size(a1), C2 = Size(a2);
		long long D1 = num[group[a1]], D2 = num[group[a2]];
		unite(a1, a2); x[cnts - 1].push_back(D1); x[cnts - 1].push_back(D2);
		p[D1] += a3*C2; p[D2] += a3*C1;
	}
	for (int i = cnts; i >= 0; i--) {
		for (int j = 0; j < x[i].size(); j++)p[x[i][j]] += p[i];
	}
	for (int i = 1; i <= n; i++) {
		cout << p[i] << endl;
	}
	return 0;
}