// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
using str = string;
int main() {
cin.tie(0)->sync_with_stdio(0);
map<str, int> IDX;
int cur = 0;
auto idx = [&](const str& S) {
if (IDX.find(S) != IDX.end()) return IDX[S];
return IDX[S] = cur++;
};
int N; cin >> N;
vector<int> P(N, -1), in(N);
vector<vector<int>> chd(N);
for(int i = 0; i < N; i++) {
str S, T; cin >> S >> T;
int s = idx(S), t = idx(T);
P[s] = t; in[t]++;
chd[t].push_back(s);
}
int left = 0, pairs = 0;
queue<int> q; for(int i = 0; i < N; i++) if (in[i] == 0) q.push(i);
set<int> cyc; for(int i = 0; i < N; i++) cyc.insert(i);
while(int(size(q)) > 0) {
int u = q.front(); q.pop();
if (!cyc.count(u)) continue;
cyc.erase(u);
int v = P[u];
if (v == -1) { left++; continue; }
cyc.erase(v);
for(auto x : chd[v]) {
P[x] = -1;
}
if (P[v] != -1) {
in[P[v]]--; if (in[P[v]] == 0) q.push(P[v]);
}
pairs++;
}
vector<int> vis(N);
for(auto i : cyc) {
int u = i, len = 0;
while(!vis[u]) {
vis[u] = 1;
u = P[u]; len++;
}
if (len == 2) continue;
left += len & 1; pairs += len / 2;
}
if (left % 2) cout << -1 << nl;
else cout << left + pairs << nl;
return 0;
}
/*
3
rocky scarlet
scarlet patrick
patrick rocky
8
leonard emmy
ada emmy
isaac leonard
emmy pierre
pierre bernhard
bernhard emmy
sofia karl
karl sofia
4
a c
b c
c d
d d
*/
