#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;
queue<int> p;
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);
    }
    bool flag = 0;
    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
        int i;
        for (i = tot;; i--) {
            if (to[st[i]].size() == 5) {
                t = i;
                break;
            }
            if (st[i] == o) break;
        }

        if (q.size() == 1) {  // 如果5度点只有一个
            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 {
        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;
}