/**
* Assume there is some hard route that goes from vertex
* u to vertex v. Let the node that the path from u to v and the
* furthest node from the hard route meet be node x. Use rerooting
* DP to calculate the hardest route and the # of such hardest routes
* for every node x.
*
* Time Complexity: O(n log(n))
*/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
int n;
cin >> n;
vector<vector<int>> adj(n);
for (int i = 1; i < n; i++) {
int x, y;
cin >> x >> y;
--x, --y;
adj[x].push_back(y);
adj[y].push_back(x);
}
vector<int> max_length(n), path_count(n);
function<void(int, int)> dfs = [&](int u, int p) {
/**
* Calculates the longest path from vertex u,
* and the number of such paths.
*/
max_length[u] = 0;
path_count[u] = 1;
for (int v : adj[u]) if (v != p) {
dfs(v, u);
if (max_length[u] < max_length[v] + 1) {
max_length[u] = max_length[v] + 1;
path_count[u] = path_count[v];
} else if (max_length[v] + 1 == max_length[u]) {
path_count[u] += path_count[v];
}
}
};
dfs(0, -1);
ll max_hardness = 0, hardest_path_count = 1;
function<void(int, int, ll, ll)> dfs2 = [&](int u, int p, ll parDist, ll parCnt) {
/**
* Performs the rerooting, to count the hardest
* path and the # of such paths at this vertex.
*/
vector<array<ll, 2>> paths; // {distance, count}
if (u > 0 || (int) adj[u].size() == 1) {
paths.push_back({parDist, parCnt});
}
for (int v : adj[u]) if (v != p) {
paths.push_back({max_length[v] + 1, path_count[v]});
}
sort(paths.begin(), paths.end(), greater<>());
if ((int) adj[u].size() >= 3) { // can form a nonzero hard route
/**
* Let the 3 longest path lengths be a, b, c, with a > b > c.
* The optimal hard route "hardness" is a * (b + c).
*/
ll a = paths[0][0], b = paths[1][0], c = paths[2][0];
ll cur = a * (b + c), num = 0, ties = 0;
for (auto [k, v] : paths) {
if (k == c) ties += v;
}
if (a != b && b != c) {
// case 1: all are distinct.
num = paths[1][1] * ties;
} else if (a == b && b == c) {
// case 2: all are the same.
num = ties * ties;
for (auto [k, v] : paths) {
if (k == a) num -= v * v;
}
num /= 2; // avoiding double counting
} else if (a == b) {
// case 3: first two are the same.
num = (paths[0][1] + paths[1][1]) * ties;
} else {
// case 4: last two are the same.
num = ties * ties;
for (auto [k, v] : paths) {
if (k == c) num -= v * v;
}
num /= 2; // avoiding double counting
}
if (max_hardness < cur) {
max_hardness = cur;
hardest_path_count = num;
} else if (max_hardness == cur) {
hardest_path_count += num;
}
}
// processing parent dist and parent count.
ll longest1 = 0;
ll longest2 = 0;
ll count1 = 0;
ll count2 = 0;
for (auto [k, v] : paths) {
if (k + 1 > longest1) {
swap(longest1, longest2);
swap(count1, count2);
longest1 = k + 1, count1 = v;
} else if (k + 1 == longest1) {
count1 += v;
} else if (k + 1 > longest2) {
longest2 = k + 1, count2 = v;
} else if (k + 1 == longest2) {
count2 += v;
}
}
for (int v : adj[u]) if (v != p) {
// using the best parent hardness and parent count possible.
if (max_length[v] + 2 == longest1) {
(path_count[v] == count1) ? dfs2(v, u, longest2, count2) :
dfs2(v, u, longest1, count1 - path_count[v]);
} else {
dfs2(v, u, longest1, count1);
}
}
};
dfs2(0, -1, 0, 1);
cout << max_hardness << ' ' << hardest_path_count << '\n';
}
