#include <algorithm>
#include <utility>
#include <cstdio>
#include <vector>
#define pb push_back
typedef long long ll; 
std::vector<std::pair<int, int> > adj[100005]; 
int x[100005], y[100005], w[100005], dep[100005], sz[100005], mx[100005]; 
void dfs(int u, int fa = -1)
{
	sz[u] = 1; 
	mx[u] = 0; 
	for (auto &e : adj[u])
	{
		int v = e.first; 
		if (v == fa)
			continue; 
		dep[v] = dep[u] + 1; 
		dfs(v, u); 
		sz[u] += sz[v]; 
		mx[u] = std::max(mx[u], sz[v]); 
	}
}
int main()
{
	// freopen("AGC018-D.in", "r", stdin); 
	int n; 
	scanf("%d", &n); 
	for (int i = 1; i < n; i++)
	{
		scanf("%d%d%d", x + i, y + i, w + i); 
		adj[--x[i]].pb({--y[i], w[i]}); 
		adj[y[i]].pb({x[i], w[i]}); 
	}
	dfs(0); 
	ll ans = 0; 
	for (int i = 1; i < n; i++)
	{
		int u = x[i], v = y[i];
		if (dep[u] > dep[v])
			std::swap(u, v); 
		int s = std::min(sz[v], n - sz[v]); 
		if (s << 1 == n)
			ans -= w[i]; 
		ans += 2ll * s * w[i]; 
	}
	for (int u = 0; u < n; u++)
	{
		mx[u] = std::max(mx[u], n - sz[u]); 
		if (mx[u] << 1 < n)
		{
			int mn = 1e9; 
			for (auto &e : adj[u])
				mn = std::min(mn, e.second); 
			ans -= mn; 
		}
	}
	printf("%lld\n", ans);
	return 0; 
}

### 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 <algorithm>
#include <utility>
#include <cstdio>
#include <vector>
#define pb push_back
typedef long long ll; 
std::vector<std::pair<int, int> > adj[100005]; 
int x[100005], y[100005], w[100005], dep[100005], sz[100005], mx[100005]; 
void dfs(int u, int fa = -1)
{
	sz[u] = 1; 
	mx[u] = 0; 
	for (auto &e : adj[u])
	{
		int v = e.first; 
		if (v == fa)
			continue; 
		dep[v] = dep[u] + 1; 
		dfs(v, u); 
		sz[u] += sz[v]; 
		mx[u] = std::max(mx[u], sz[v]); 
	}
}
int main()
{
	// freopen("AGC018-D.in", "r", stdin); 
	int n; 
	scanf("%d", &n); 
	for (int i = 1; i < n; i++)
	{
		scanf("%d%d%d", x + i, y + i, w + i); 
		adj[--x[i]].pb({--y[i], w[i]}); 
		adj[y[i]].pb({x[i], w[i]}); 
	}
	dfs(0); 
	ll ans = 0; 
	for (int i = 1; i < n; i++)
	{
		int u = x[i], v = y[i];
		if (dep[u] > dep[v])
			std::swap(u, v); 
		int s = std::min(sz[v], n - sz[v]); 
		if (s << 1 == n)
			ans -= w[i]; 
		ans += 2ll * s * w[i]; 
	}
	for (int u = 0; u < n; u++)
	{
		mx[u] = std::max(mx[u], n - sz[u]); 
		if (mx[u] << 1 < n)
		{
			int mn = 1e9; 
			for (auto &e : adj[u])
				mn = std::min(mn, e.second); 
			ans -= mn; 
		}
	}
	printf("%lld\n", ans);
	return 0; 
}

```