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 <bits/stdc++.h>
using namespace std;

typedef long long ll;

const int MAXN = 1e5 + 5;

struct edge {
	int a, b, c;
	bool operator <(const edge &rhs) const {
		return c > rhs.c;
	}
};

int N;
vector <edge> E;
ll sol[MAXN];
vector <int> comp[MAXN];
int dad[MAXN];
ll add[MAXN];

void load() {
	scanf("%d", &N);
	for (int i = 1; i < N; i++) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		E.push_back({a, b, c});
	}
}

void join(edge e) {
	int da = dad[e.a];
	int db = dad[e.b];
	int sza = comp[da].size();
	int szb = comp[db].size();
	if (sza > szb) {
		swap(da, db);
		swap(sza, szb);
	}
	for (auto it : comp[da]) {
		sol[it] += add[da] - add[db] + (ll)(szb - sza) * e.c;
		comp[db].push_back(it);
		dad[it] = db;
	}
	add[db] += (ll)sza * e.c;
}

void solve() {
	for (int i = 1; i <= N; i++) {
		dad[i] = i;
		comp[i].push_back(i);
	}
	
	sort(E.begin(), E.end());
	for (auto it : E)
		join(it);
	
	for (int i = 1; i <= N; i++)
		printf("%lld\n", sol[i] + add[dad[i]]);
}

int main() {
	load();
	solve();
	return 0;
}