#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

struct E{ int x, c; };

int n, s[100010], c, ms;
ll ans, msd, md = 1e9;
vector<E> e[100010];

int sf(int x, int p){
	s[x] = 1;
	for(auto &i : e[x]){
		if(i.x == p) continue;
		s[x] += sf(i.x, x);
	}
	return s[x];
}

int cf(int x, int p, int t){
	for(auto &i : e[x]){
		if(i.x != p && s[i.x] > t / 2) return cf(i.x, x, t);
	}
	return x;
}

void g(int x, int p, ll d){
	ans += 2 * d;
	for(auto &i : e[x]){
		if(i.x == p) continue;
		if(ms < s[i.x]){
			ms = s[i.x];
			msd = d + i.c;
		}
		md = min(md, d + i.c);
		g(i.x, x, d + i.c);
	}
}

int main(){
	scanf("%d", &n);
	for(int i = 0, x, y, z; i < n - 1; i++){
		scanf("%d%d%d", &x, &y, &z);
		e[x].push_back({y, z});
		e[y].push_back({x, z});
	}
	c = cf(1, 0, sf(1, 0));
	sf(c, 0);
	g(c, 0, 0);
	if(!(n & 1) && ms == n / 2) printf("%lld\n", ans - msd);
	else printf("%lld\n", ans - md);
}
### Prompt

Your challenge is to write a CPP solution to the following problem:

There is a tree with N vertices, numbered 1 through N. The i-th edge in this tree connects Vertices A_i and B_i and has a length of C_i.

Joisino created a complete graph with N vertices. The length of the edge connecting Vertices u and v in this graph, is equal to the shortest distance between Vertices u and v in the tree above.

Joisino would like to know the length of the longest Hamiltonian path (see Notes) in this complete graph. Find the length of that path.

Constraints

* 2 \leq N \leq 10^5
* 1 \leq A_i < B_i \leq N
* The given graph is a tree.
* 1 \leq C_i \leq 10^8
* All input values are integers.

Input

Input is given from Standard Input in the following format:


N
A_1 B_1 C_1
A_2 B_2 C_2
:
A_{N-1} B_{N-1} C_{N-1}


Output

Print the length of the longest Hamiltonian path in the complete graph created by Joisino.

Examples

Input

5
1 2 5
3 4 7
2 3 3
2 5 2


Output

38


Input

8
2 8 8
1 5 1
4 8 2
2 5 4
3 8 6
6 8 9
2 7 12


Output

132

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

struct E{ int x, c; };

int n, s[100010], c, ms;
ll ans, msd, md = 1e9;
vector<E> e[100010];

int sf(int x, int p){
	s[x] = 1;
	for(auto &i : e[x]){
		if(i.x == p) continue;
		s[x] += sf(i.x, x);
	}
	return s[x];
}

int cf(int x, int p, int t){
	for(auto &i : e[x]){
		if(i.x != p && s[i.x] > t / 2) return cf(i.x, x, t);
	}
	return x;
}

void g(int x, int p, ll d){
	ans += 2 * d;
	for(auto &i : e[x]){
		if(i.x == p) continue;
		if(ms < s[i.x]){
			ms = s[i.x];
			msd = d + i.c;
		}
		md = min(md, d + i.c);
		g(i.x, x, d + i.c);
	}
}

int main(){
	scanf("%d", &n);
	for(int i = 0, x, y, z; i < n - 1; i++){
		scanf("%d%d%d", &x, &y, &z);
		e[x].push_back({y, z});
		e[y].push_back({x, z});
	}
	c = cf(1, 0, sf(1, 0));
	sf(c, 0);
	g(c, 0, 0);
	if(!(n & 1) && ms == n / 2) printf("%lld\n", ans - msd);
	else printf("%lld\n", ans - md);
}
```