#include<bits/stdc++.h>
using namespace std;
int n, mx[100007];
long long res;
vector<pair<int, int> >e[100007];
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int dfs(int w, int f, int v)
{
	int sz = 1;
	for (int i = e[w].size() - 1; i >= 0; i--)
		if (e[w][i].first != f)
		{
			int s = dfs(e[w][i].first, w, e[w][i].second);
			mx[w] = max(mx[w], s);
			sz += s;
		}
	mx[w] = max(mx[w], n - sz);
	res += 2LL * v * min(sz, n - sz);
	return sz;
}
int main()
{
	n = read();
	for (int i = 2; i <= n; i++)
	{
		int a = read(), b = read(), c = read();
		e[a].push_back(make_pair(b, c));
		e[b].push_back(make_pair(a, c));
	}
	dfs(1, 0, 0);
	for (int i = 1; i <= n; i++)if (mx[i] <= (n / 2))
		for (int j = i + 1; j <= n; j++)if (mx[j] <= (n / 2))
			for (int k = e[i].size() - 1; k >= 0; k--)
				if (e[i][k].first == j)
				{
					cout << res - e[i][k].second << endl;
					return 0;
				}
	int mn = 1e9;
	for (int i = 1; i <= n; i++)if (mx[i] <= (n / 2))
	{
		for (int k = e[i].size() - 1; k >= 0; k--)
			mn = min(mn, e[i][k].second);
	}
	cout << res - mn << endl;
	return 0;
}
### Prompt

Please formulate 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;
int n, mx[100007];
long long res;
vector<pair<int, int> >e[100007];
int read()
{
	int num = 0;
	char c = getchar();
	while (c < '0' || c>'9')c = getchar();
	while (c >= '0' && c <= '9')num = num * 10 + c - '0', c = getchar();
	return num;
}
int dfs(int w, int f, int v)
{
	int sz = 1;
	for (int i = e[w].size() - 1; i >= 0; i--)
		if (e[w][i].first != f)
		{
			int s = dfs(e[w][i].first, w, e[w][i].second);
			mx[w] = max(mx[w], s);
			sz += s;
		}
	mx[w] = max(mx[w], n - sz);
	res += 2LL * v * min(sz, n - sz);
	return sz;
}
int main()
{
	n = read();
	for (int i = 2; i <= n; i++)
	{
		int a = read(), b = read(), c = read();
		e[a].push_back(make_pair(b, c));
		e[b].push_back(make_pair(a, c));
	}
	dfs(1, 0, 0);
	for (int i = 1; i <= n; i++)if (mx[i] <= (n / 2))
		for (int j = i + 1; j <= n; j++)if (mx[j] <= (n / 2))
			for (int k = e[i].size() - 1; k >= 0; k--)
				if (e[i][k].first == j)
				{
					cout << res - e[i][k].second << endl;
					return 0;
				}
	int mn = 1e9;
	for (int i = 1; i <= n; i++)if (mx[i] <= (n / 2))
	{
		for (int k = e[i].size() - 1; k >= 0; k--)
			mn = min(mn, e[i][k].second);
	}
	cout << res - mn << endl;
	return 0;
}
```