#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair <ll, ll> pll;

vector <pll> V[101010];
ll S[101010], dp[101010][3];
ll n;

void dfs(ll p, ll r)
{
	ll s0, s1f, s1s, s2;
	ll k0, k1, k2;
	
	S[p] = 1;
	s0 = 0;
	s1f = s1s = s2 = -1e16;
	
	for(pll t: V[p]){
		if(t.first != r){
			dfs(t.first, p);
			S[p] += S[t.first];
			k0 = dp[t.first][0] + t.second * (min(S[t.first], (n - S[t.first]) - 1) * 2);
			k1 = dp[t.first][1] + t.second * (min(S[t.first], (n - S[t.first])) * 2 - 1);
			k2 = dp[t.first][2] + t.second * (min(S[t.first] - 1, (n - S[t.first])) * 2);
			s0 += k0;
			if(s1f < k1 - k0) s1s = s1f, s1f = k1 - k0;
			else if(s1s < k1 - k0) s1s = k1 - k0;
			s2 = max(s2, k2 - k0);
		}
	}
	
	dp[p][0] = s0;
	dp[p][1] = max(s0, s0 + s1f);
	dp[p][2] = max(max(s0 + s2, s0 + s1f + s1s), s0 + s1f);
}

int main()
{
	ll i, a, b, c;
	
	scanf("%lld", &n);
	
	for(i=1;i<n;i++){
		scanf("%lld%lld%lld", &a, &b, &c);
		V[a].push_back(pll(b, c));
		V[b].push_back(pll(a, c));
	}
	
	dfs(1, 0);
	
	printf("%lld\n", dp[1][2]);
	
	return 0;
}
### Prompt

In CPP, your task is to solve 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;
typedef pair <ll, ll> pll;

vector <pll> V[101010];
ll S[101010], dp[101010][3];
ll n;

void dfs(ll p, ll r)
{
	ll s0, s1f, s1s, s2;
	ll k0, k1, k2;
	
	S[p] = 1;
	s0 = 0;
	s1f = s1s = s2 = -1e16;
	
	for(pll t: V[p]){
		if(t.first != r){
			dfs(t.first, p);
			S[p] += S[t.first];
			k0 = dp[t.first][0] + t.second * (min(S[t.first], (n - S[t.first]) - 1) * 2);
			k1 = dp[t.first][1] + t.second * (min(S[t.first], (n - S[t.first])) * 2 - 1);
			k2 = dp[t.first][2] + t.second * (min(S[t.first] - 1, (n - S[t.first])) * 2);
			s0 += k0;
			if(s1f < k1 - k0) s1s = s1f, s1f = k1 - k0;
			else if(s1s < k1 - k0) s1s = k1 - k0;
			s2 = max(s2, k2 - k0);
		}
	}
	
	dp[p][0] = s0;
	dp[p][1] = max(s0, s0 + s1f);
	dp[p][2] = max(max(s0 + s2, s0 + s1f + s1s), s0 + s1f);
}

int main()
{
	ll i, a, b, c;
	
	scanf("%lld", &n);
	
	for(i=1;i<n;i++){
		scanf("%lld%lld%lld", &a, &b, &c);
		V[a].push_back(pll(b, c));
		V[b].push_back(pll(a, c));
	}
	
	dfs(1, 0);
	
	printf("%lld\n", dp[1][2]);
	
	return 0;
}
```