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;

const int N = 100005;

int n;
int par[N], sz[N];
vector <int> vec[N];
long long res[N];
long long lz[N];

struct edge {
	int w; int u; int v;
	bool operator < (const edge &other) const {
		return w > other.w;
	}
};
vector <edge> ed;

int anc(int p) { return p == par[p] ? p : par[p] = anc(par[p]); }

void join(int p, int q, int w) {
	p = anc(p), q = anc(q);
	if (vec[p].size() > vec[q].size()) swap(p, q);

	while(vec[p].size()) {
		int u = vec[p].back();
		res[u] += lz[p];
		res[u] -= lz[q];
		res[u] += 1LL * (sz[q] - sz[p]) * w;
		vec[q].push_back(u);
		vec[p].pop_back();
	}

	par[p] = q;
	lz[q] += 1LL * sz[p] * w;
	sz[q] += sz[p];
	lz[p] = 0;
	sz[p] = 0;
}

int main() {
	ios_base::sync_with_stdio(false); cin.tie(0);
	
	cin >> n;
	for (int i = 1; i < n; ++i) {
		int u, v, w; cin >> u >> v >> w;
		ed.push_back({w, u, v});
	}
	sort(ed.begin(), ed.end());

	for (int i = 1; i <= n; ++i) vec[i].push_back(i), par[i] = i, sz[i] = 1;

	for (auto &e : ed) {
		join(e.u, e.v, e.w);
	}

	for (int i = 1; i <= n; ++i) if (i == anc(i)) {
		for (auto &j : vec[i]) {
			res[j] += lz[i];
		}
	}

	for (int i = 1; i <= n; ++i) printf("%lld\n", res[i]);
}