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#include <algorithm>
#include <iostream>
#include <set>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;
const int LIM = 800008;
int N;
int A[LIM], B[LIM];
void solve() {
vector<tuple<int, bool, int>> E;
set<pair<int, int>> S;
// Input, while finding tallest grounded ladder.
int ng = 0, mg = -1;
cin >> N;
for (int i = 0; i < N; i++) {
cin >> A[i] >> B[i];
if (!A[i]) {
ng++;
if (mg < 0 || B[i] > B[mg]) {
mg = i;
}
} else {
E.emplace_back(A[i], 0, i);
E.emplace_back(B[i], 1, i);
}
}
// None grounded?
if (ng == 0) {
cout << "0 0" << endl;
return;
}
// Sort endpoints (including tallest grounded ladder).
E.emplace_back(B[mg], 1, mg);
sort(E.begin(), E.end());
// Line sweep.
int ans1 = ng, ans2 = 0;
int a = mg, b = mg;
S.clear();
for (auto e : E) {
int i = get<2>(e);
if (!get<1>(e)) {
// Bottom of ladder i.
S.emplace(B[i], i);
continue;
}
// Top of ladder i.
if (B[a] > B[b]) {
swap(a, b);
}
auto it = S.find(make_pair(B[i], i));
if (it != S.end()) {
// Ladder i not used yet, so jump from ladder a to it.
S.erase(it);
ans1++;
ans2++;
a = i;
} else if (i == b) {
// Consider ladder i to be ladder a rather than b.
swap(a, b);
} else if (i != a) {
// Ladder i is irrelevant.
continue;
}
// Top of ladder a, so need to jump off of it.
if (!S.empty()) {
// There are other unused ladders, so jump from ladder a to the one with
// lowest top.
bool isB = a == b;
a = S.begin()->second;
S.erase(make_pair(B[a], a));
ans1++;
ans2++;
if (isB) {
// This is also ladder b, so repeat for it.
if (!S.empty()) {
// Jump from ladder b to another unused ladder.
b = S.begin()->second;
S.erase(make_pair(B[b], b));
ans1++;
ans2++;
} else {
// Jump from ladder b to the new ladder a.
ans2++;
b = a;
}
}
} else if (a != b) {
// Jump from ladder a to ladder b.
ans2++;
a = b;
} else {
// Nowhere else to go, so stop.
break;
}
}
cout << ans1 << " " << ans2 << endl;
}
int main() {
int T;
cin >> T;
for (int t = 1; t <= T; t++) {
cout << "Case #" << t << ": ";
solve();
}
return 0;
}
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