/*
 A tree needs more than 10000 nodes to use the 7th color */
#include <stdio.h>
#include <string.h>
#include <vector>

using namespace std;
const int N = 1e4+5;
const int M = 6;
const int INF = 0x3f3f3f3f;

int n, dp[N][M+5];
vector<int> g[N];

bool get () {

    char s[N*M];
    gets(s);

    if (s[0] == '\0') return false;

    int len = strlen(s), i = 0;

    int u = -1, v = -1;
    while (s[i] < '0' && s[i] > '9') i++;

    for (; s[i] != ':'; i++) {
        if (u == -1) u = s[i] - '0';
        else u = u * 10 + s[i] - '0';
    }

    for (; i <= len; i++) {
        if (s[i] >= '0' && s[i] <= '9') {
            if (v == -1) v = s[i] - '0';
            else v = v * 10 + s[i] - '0';
        } else {
            if (v != -1) {
                g[u].push_back(v);
                g[v].push_back(u);
                v = -1;
            }
        }
    }

    return true;
}

void dfs (int u, int f) { ///(cur_node,prev_node)
    for (int i = 1; i <= M; i++)
        dp[u][i] = i;

    for (int i = 0; i < g[u].size(); i++) {

        int v = g[u][i];
        if (v == f) continue;

        dfs(v, u);
 /**Basically since x,y are colors, the condition if(x==y) signify that no two connected nodes can
have same color, and what we are doing in the nested for loops is that if we choose to color parent
node with 'x' then what is the minimum color we can assign to the children is the cur, so
dp[u][x]+=cur (i.e. dp[u][x] is containing the minimum sum of colors of the parent and all its children when the parent u
is colored x)**/
        for (int x = 1; x <= M; x++) {
            int cur = INF;
            for (int y = 1; y <= M; y++) {
                if (x == y) continue;
                cur = min(cur, dp[v][y]);
            }
            dp[u][x] += cur;
        }
    }
}

int main () {
    while (scanf("%d%*c", &n), n) {
        for (int i = 0; i < n; i++)
            g[i].clear();

        while (get()) ;

        dfs(0, -1);
        int ans = INF;
        for (int i = 1; i <= M; i++)
            ans = min(ans, dp[0][i]);
        printf("%d\n", ans);
    }
    return 0;
}