#include <bits/stdc++.h>
using namespace std;
#define maxn 200005
int n;
int x, y, cnt, st[maxn], tot, o, t;
int ans = 0;
vector<int> to[maxn];
bool flat[maxn];
bool dfs(int x, int fa)
{
    st[++tot] = x;
    flat[x] = 1;
    for (int i = 0; i < to[x].size(); i++)
    {
        if (fa == to[x][i])
            continue;
        if (flat[to[x][i]])
        {
            o = to[x][i];
            return 1;
        }
        else if (dfs(to[x][i], x))
        {
            return 1;
        }
    }
    --tot;
    return 0;
}
queue<int> q;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> x >> y;
        to[x].push_back(y);
        to[y].push_back(x);
    }
    for (int i = 1; i <= n; i++)
    {
        if (to[i].size() == 5)
        {
            q.push(i);
        }
        else
        {
            if (to[i].size() != 4)
                cnt++;
        }
    }
    dfs(1, 0);

    if (!q.empty())
    { // 有5度的点的话，删除的边必须包含所有的该点

        if (q.size() == 1)
        { // 如果5度点只有一个
            for (int i = tot;; i--)
            {
                if (to[st[i]].size() == 5)
                {
                    t = i;
                    break;
                }
            }
            if (st[t] == o)
                st[t - 1] = st[tot];
            if (t == tot)
                st[t + 1] = o;
            if (to[st[t + 1]].size() == 4)
                ans++;
            if (to[st[t - 1]].size() == 4)
                ans++;
            cout << ans + cnt * 2 << endl;
            return 0;
        }
        else
        { // 5度点有两个
            cout << cnt << endl;
            return 0;
        }
    }
    else
    {//可以删除任意一条环里面的边

        for (int i = tot; st[i] != o; i--)
        { // 枚举的是删去 st[i] - st[i-1] 这条边
            if (to[st[i]].size() == 4)
                ans++;
            if (to[st[i - 1]].size() == 4)
                ans++;
            ans += cnt;
        }

        if (to[st[tot]].size() == 4)
            ans++;
        if (to[o].size() == 4)
            ans++;
        ans += cnt;

        cout << ans << endl;
    }
    return 0;
}