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<cstdio>
#include<algorithm>
#include<vector>
#include<deque>
#include<stack>
#include<queue>
#include<string>
#include<iostream>
#include<tuple>
#include<utility>
#include<set>
#include<queue>
#include<iomanip>
#include<iterator>
#include<map>
#include<bitset>
#include<array>
#include<cmath>
#include<functional>
#include<list>
using namespace std;
typedef long long int llint;
typedef long double lldo;
const llint big=2e18+100000;
const int mod=1e9+7;
const lldo eps=1e-7;
const long double pai=3.141592653589793238462643383279502884197;
#define mt make_tuple
#define mp make_pair
#define fir first
#define sec second
#define pub push_back
#define puf push_front
#define pob pop_back
#define pof pop_front
#define res resize
#define ins insert
#define era erase
#define dme(in) cout<<in<<endl;return 0
template <class T,class U>void mineq(T& a,U b){if(a>b){a=b;}}
template <class T,class U>void maxeq(T& a,U b){if(a<b){a=b;}}
int main(void){
	int i,n;cin>>n;
	vector<tuple<llint,int,int>>hen(n-1);
	vector<int>oya(n);
	vector<int>siz(n,1);
	vector<llint>imo(n);//下方向
	for(i=0;i<n;i++){oya[i]=i;}
	for(i=0;i<n-1;i++){
		int a,b,c;cin>>a>>b>>c;a--;b--;
		hen[i]=mt(c,a,b);
	}
	sort(hen.rbegin(),hen.rend());
	//すぬけ 木である必要はない
	for(i=0;i<n-1;i++){
		int a,b;llint c;
		tie(c,a,b)=hen[i];
		while(a!=oya[a]){a=oya[a];}
		while(b!=oya[b]){b=oya[b];}
		if(a==b){continue;}
		if(siz[a]<siz[b]){swap(a,b);}
		imo[a]+=c*siz[b];
		imo[b]+=c*siz[a]-imo[a];
		siz[a]+=siz[b];
		oya[b]=a;
	}
	vector<pair<int,int>>miru(n);
	for(i=0;i<n;i++){miru[i]=mp(siz[i],i);}
	//子のほうがサイズが小さいので親から見る=サイズの大きい順
	sort(miru.rbegin(),miru.rend());
	for(i=1;i<n;i++){//最大は根なので
		imo[miru[i].sec]+=imo[oya[miru[i].sec]];
	}
	for(i=0;i<n;i++){cout<<imo[i]<<endl;}
	return 0;
}