//#include <tourist>
#include <bits/stdc++.h>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<ll, ll> p;
const int INF = 1e9;
const ll LINF = ll(1e18);
const int MOD = 1000000007;
const int dx[4] = {0, 1, 0, -1}, dy[4] = {-1, 0, 1, 0};
const int Dx[8] = {0, 1, 1, 1, 0, -1, -1, -1}, Dy[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
#define yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define no cout << "No" << endl
#define NO cout << "NO" << endl
#define rep(i, n) for (int i = 0; i < n; i++)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define ALL(v) v.begin(), v.end()
#define debug(v)          \
    cout << #v << ":";    \
    for (auto x : v)      \
    {                     \
        cout << x << ' '; \
    }                     \
    cout << endl;
template <class T>
bool chmax(T &a, const T &b)
{
    if (a < b)
    {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T &a, const T &b)
{
    if (b < a)
    {
        a = b;
        return 1;
    }
    return 0;
}
//cout<<fixed<<setprecision(15);有効数字15桁
//-std=c++14
//g++ yarudake.cpp -std=c++17 -I .
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
ll dfs(ll k,vector<vector<pair<ll,ll>>> &v,vector<ll> &count,ll prev){
    ll res=1;
    rep(i,v[k].size()){
        if(v[k][i].first==prev)continue;
        res+=dfs(v[k][i].first,v,count,k);
    }
    count[k]=res;
    return res;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll n;
    cin>>n;
    vector<vector<pair<ll,ll>>> v(n);
    rep(i,n-1){
        ll a,b,c;
        cin>>a>>b>>c;
        a--;
        b--;
        v[a].push_back(make_pair(b,c));
        v[b].push_back(make_pair(a, c));
    }
    vector<ll> count(n);
    dfs(0,v,count,-1);
    ll ans=0;
    //debug(count);
    rep(i,n){
        rep(j,v[i].size()){
            //if(i>v[i][j].first)continue;
            ans += v[i][j].second * min(count[i], count[v[i][j].first]) * (n - min(count[i], count[v[i][j].first]));
        }
    }
    cout<<ans<<"\n";
}