#include <bits/stdc++.h>
using namespace std;
int n;
vector<vector<int>> adj;
vector<int> child, height;
void dfs_child(int u, int p = -1) {
  for (int v : adj[u])
    if (v ^ p) {
      dfs_child(v, u);
      child[u] += child[v] + 1;
    }
}
void dfs_height(int u, int p = -1) {
  if (p >= 0) height[u] = height[p] + 1;
  for (int v : adj[u])
    if (v ^ p) dfs_height(v, u);
}
int dfs_max(int u = 0, int p = -1) {
  int cnt = 0;
  int add = 0;
  for (int v : adj[u])
    if (v ^ p) {
      if (child[v] == 0)
        add = 1;
      else
        cnt++;
      cnt += dfs_max(v, u);
    }
  return cnt + add;
}
int calcMin() {
  set<int> mod;
  for (int i = 0; i < n; i++)
    if (child[i] == 0) mod.insert(height[i] % 2);
  if (mod.size() == 1) return 1;
  return 3;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> n;
  adj.resize(n);
  child.resize(n, 0);
  height.resize(n, 0);
  for (int i = 0, u, v; i < n - 1; i++) {
    cin >> u >> v, u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int p = 0;
  for (int u = 0; u < n; u++)
    if (adj[u].size() != 1) {
      p = u;
      break;
    }
  dfs_child(p);
  dfs_height(p);
  cout << calcMin() << " " << dfs_max(p) << endl;
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

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;
int n;
vector<vector<int>> adj;
vector<int> child, height;
void dfs_child(int u, int p = -1) {
  for (int v : adj[u])
    if (v ^ p) {
      dfs_child(v, u);
      child[u] += child[v] + 1;
    }
}
void dfs_height(int u, int p = -1) {
  if (p >= 0) height[u] = height[p] + 1;
  for (int v : adj[u])
    if (v ^ p) dfs_height(v, u);
}
int dfs_max(int u = 0, int p = -1) {
  int cnt = 0;
  int add = 0;
  for (int v : adj[u])
    if (v ^ p) {
      if (child[v] == 0)
        add = 1;
      else
        cnt++;
      cnt += dfs_max(v, u);
    }
  return cnt + add;
}
int calcMin() {
  set<int> mod;
  for (int i = 0; i < n; i++)
    if (child[i] == 0) mod.insert(height[i] % 2);
  if (mod.size() == 1) return 1;
  return 3;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr), cout.tie(nullptr);
  cin >> n;
  adj.resize(n);
  child.resize(n, 0);
  height.resize(n, 0);
  for (int i = 0, u, v; i < n - 1; i++) {
    cin >> u >> v, u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int p = 0;
  for (int u = 0; u < n; u++)
    if (adj[u].size() != 1) {
      p = u;
      break;
    }
  dfs_child(p);
  dfs_height(p);
  cout << calcMin() << " " << dfs_max(p) << endl;
  return 0;
}

```