/*
思路:
可以将城市与首都之间抽象成一个无环连通图，即为一棵树
然后找出树的直径;
*/
#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+10;
int h[N];
const int M = 2*N;
int e[M],w[M],ne[M],idx;
int n;
int dist[N];
void dfs(int u,int f,int d)
{
    dist[u] = d; 
    for(int i = h[u];i != -1; i = ne[i])
    {
        int j = e[i];
        if(j != f) dfs(j,u,d+w[i]);
    }
}

void add(int a,int b,int c)
{
    w[idx] = c, e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}

int main()
{
    cin >> n;
    memset(h,-1,sizeof h);
    for(int i =0 ; i<n ;i++)
    {
        int a,b ,w ;cin >> a >> b >>w;
        add(a,b,w); add(b,a,w);
    }

    dfs(1,-1,0);  //第一次找最远点
    int zyd = 0;
    for(int i =1; i <= n ;i++)
    {
        if(dist[zyd] < dist[i]) zyd = i;
    }

    memset(dist,0,sizeof dist); //第二次找最远点
    dfs(zyd,-1,0);
    zyd = 0;
    for(int i =1; i <= n ;i++)
    {
        if(dist[zyd] < dist[i]) zyd = i;
    }

    long long res = dist[zyd] * 10 + (dist[zyd] + 1ll) * dist[zyd] / 2;
    cout << res;
}