#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 2;
const int mod = 1e9 + 7;
vector<int> G[N];
int degree[N];
int ansMx, ansMn;
pair<int, int> dfs(int u, int p, int fl) {
  int leaves = 0;
  pair<int, int> cur = {0, 0};
  if (G[u].size() == 1) {
    if (fl) return {0, 1};
    return {1, 0};
  }
  for (auto v : G[u]) {
    if (v == p) continue;
    pair<int, int> go = dfs(v, u, 1 - fl);
    cur.first |= go.first;
    cur.second |= go.second;
    if (G[v].size() == 1) leaves++;
  }
  if (leaves) ansMx = ansMx - leaves + 1;
  if (cur.first && cur.second) ansMn = 3;
  return cur;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  ansMx = n - 1;
  ansMn = 1;
  int root = 0;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
    degree[u]++;
    degree[v]++;
    if (degree[u] > 1) root = u;
    if (degree[v] > 1) root = v;
  }
  dfs(root, 0, 0);
  cout << ansMn << ' ' << ansMx << endl;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

You have unweighted tree of n vertices. You have to assign a positive weight to each edge so that the following condition would hold:

  * For every two different leaves v_{1} and v_{2} of this tree, [bitwise XOR](https://en.wikipedia.org/wiki/Bitwise_operation#XOR) of weights of all edges on the simple path between v_{1} and v_{2} has to be equal to 0. 



Note that you can put very large positive integers (like 10^{(10^{10})}).

It's guaranteed that such assignment always exists under given constraints. Now let's define f as the number of distinct weights in assignment.

<image> In this example, assignment is valid, because bitwise XOR of all edge weights between every pair of leaves is 0. f value is 2 here, because there are 2 distinct edge weights(4 and 5).

<image> In this example, assignment is invalid, because bitwise XOR of all edge weights between vertex 1 and vertex 6 (3, 4, 5, 4) is not 0. 

What are the minimum and the maximum possible values of f for the given tree? Find and print both.

Input

The first line contains integer n (3 ≤ n ≤ 10^{5}) — the number of vertices in given tree.

The i-th of the next n-1 lines contains two integers a_{i} and b_{i} (1 ≤ a_{i} < b_{i} ≤ n) — it means there is an edge between a_{i} and b_{i}. It is guaranteed that given graph forms tree of n vertices.

Output

Print two integers — the minimum and maximum possible value of f can be made from valid assignment of given tree. Note that it's always possible to make an assignment under given constraints.

Examples

Input


6
1 3
2 3
3 4
4 5
5 6


Output


1 4


Input


6
1 3
2 3
3 4
4 5
4 6


Output


3 3


Input


7
1 2
2 7
3 4
4 7
5 6
6 7


Output


1 6

Note

In the first example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>

In the second example, possible assignments for each minimum and maximum are described in picture below. The f value of valid assignment of this tree is always 3. 

<image>

In the third example, possible assignments for each minimum and maximum are described in picture below. Of course, there are multiple possible assignments for each minimum and maximum. 

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 2;
const int mod = 1e9 + 7;
vector<int> G[N];
int degree[N];
int ansMx, ansMn;
pair<int, int> dfs(int u, int p, int fl) {
  int leaves = 0;
  pair<int, int> cur = {0, 0};
  if (G[u].size() == 1) {
    if (fl) return {0, 1};
    return {1, 0};
  }
  for (auto v : G[u]) {
    if (v == p) continue;
    pair<int, int> go = dfs(v, u, 1 - fl);
    cur.first |= go.first;
    cur.second |= go.second;
    if (G[v].size() == 1) leaves++;
  }
  if (leaves) ansMx = ansMx - leaves + 1;
  if (cur.first && cur.second) ansMn = 3;
  return cur;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  ansMx = n - 1;
  ansMn = 1;
  int root = 0;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
    degree[u]++;
    degree[v]++;
    if (degree[u] > 1) root = u;
    if (degree[v] > 1) root = v;
  }
  dfs(root, 0, 0);
  cout << ansMn << ' ' << ansMx << endl;
}

```