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;
#define rep(i,n) for(int i=0; i<(n); i++)
constexpr int INF = 1e9+7, MX = 100005;
using T = tuple<int,int,int>;
int p[MX];
long long ans[MX], laz[MX];
vector<int> ch[MX];
T ts[MX];
int n, a, b, c;

int main(){
	scanf("%d", &n);
	rep(i,n-1){
		scanf("%d%d%d", &a, &b, &c);
		ts[i] = T{c,a-1,b-1};
	}
	sort(ts, ts+n-1, greater<T>());
	rep(i,n) ch[i].emplace_back(p[i] = i);
	rep(i,n-1){
		tie(c, a, b) = ts[i];
		a = p[a]; b = p[b];
		if(ch[a].size() < ch[b].size()) swap(a,b);
		long long sd = ch[a].size() - ch[b].size();
		for(int d: ch[b]){
			p[d] = a;
			ans[d] += laz[b] + sd * c - laz[a];
			ch[a].emplace_back(d);
		}
		laz[a] += (long long) ch[b].size() * c;
	}
	rep(i,n) printf("%lld\n", ans[i] + laz[p[0]]);
}