#include <iostream>
#include <cstring>

using namespace std;

const int N = 110, M = 2 * N, INF = 1e9;

int n;
int cnt[N];                             //cnt用来存每个村的人口
int h[N], e[M], ne[M], idx;             //邻接表

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

int dfs(int u, int father, int dist)    //当前为u节点，u节点从哪个节点来，当前节点与根节点的距离
{
    int sum = cnt[u] * dist;            //当前点的值 = 人数 * 与根节点的距离

    for (int i = h[u]; ~i; i = ne[i])
    {
        int j = e[i];
        if (j == father) continue;      //不走回头路
        sum += dfs(j, u, dist + 1);     //下一节点为j，从u过去，距离加1
    }

    return sum;
}

int main()
{
    cin >> n;

    memset(h, -1, sizeof h);

    for (int i = 1; i <= n; i ++ )
    {
        int l, r;
        cin >> cnt[i] >> l >> r;        //读入每个节点人数、左儿子和右儿子

        if (l) add(i, l), add(l, i);    //儿子节点不为空才建立无向边
        if (r) add(i, r), add(r, i);
    }

    int res = INF;                      //遍历找出最小值
    for (int i = 1; i <= n; i ++ ) res = min(res, dfs(i, -1, 0));

    cout << res << endl;

    return 0;
}