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

typedef long long ll;
typedef pair<int, int> pii;
#define x first
#define y second
const int maxn = 1e5 + 10;
int n, sz[maxn], dpar[maxn], mn = 1e9;
vector<pii> adj[maxn];
vector<int> cent;
ll tot;

void dfs(int i, int p){
    sz[i] = 1;
    int m = 0;

    for(pii j : adj[i]){
        if(j.x != p){
            dpar[j.x] = j.y;
            dfs(j.x, i);
            sz[i] += sz[j.x];
            m = max(m, sz[j.x]);
        }
    }

    m = max(m, n - sz[i]);

    if(m <= n / 2){
        cent.push_back(i);
    }

    if(p != 0){
        tot += ll(dpar[i]) * 2 * min(sz[i], n - sz[i]);
    }
}

int main(){
    ios_base::sync_with_stdio(false);

    cin >> n;

    for(int i = 1; i < n; i++){
        int a, b, c;
        cin >> a >> b >> c;
        adj[a].push_back({b, c});
        adj[b].push_back({a, c});
    }

    dfs(1, 0);

    if(cent.size() == 1){
        for(pii j : adj[cent[0]]){
            mn = min(mn, j.y);
        }
    }else{
        for(pii j : adj[cent[0]]){
            if(j.x == cent[1]){
                mn = min(mn, j.y);
            }
        }
    }

    cout << tot - mn << endl;

    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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<int, int> pii;
#define x first
#define y second
const int maxn = 1e5 + 10;
int n, sz[maxn], dpar[maxn], mn = 1e9;
vector<pii> adj[maxn];
vector<int> cent;
ll tot;

void dfs(int i, int p){
    sz[i] = 1;
    int m = 0;

    for(pii j : adj[i]){
        if(j.x != p){
            dpar[j.x] = j.y;
            dfs(j.x, i);
            sz[i] += sz[j.x];
            m = max(m, sz[j.x]);
        }
    }

    m = max(m, n - sz[i]);

    if(m <= n / 2){
        cent.push_back(i);
    }

    if(p != 0){
        tot += ll(dpar[i]) * 2 * min(sz[i], n - sz[i]);
    }
}

int main(){
    ios_base::sync_with_stdio(false);

    cin >> n;

    for(int i = 1; i < n; i++){
        int a, b, c;
        cin >> a >> b >> c;
        adj[a].push_back({b, c});
        adj[b].push_back({a, c});
    }

    dfs(1, 0);

    if(cent.size() == 1){
        for(pii j : adj[cent[0]]){
            mn = min(mn, j.y);
        }
    }else{
        for(pii j : adj[cent[0]]){
            if(j.x == cent[1]){
                mn = min(mn, j.y);
            }
        }
    }

    cout << tot - mn << endl;

    return 0;
}

```