#include <bits/stdc++.h>
using namespace std;
vector<int> adj[100005];
int dep[100005];
int minu;
int f;
void dfs(int node, int par) {
  dep[node] = 1;
  int mx = 0;
  for (int i = 0; i < adj[node].size(); i++)
    if (adj[node][i] != par) {
      dfs(adj[node][i], node);
      mx = max(mx, dep[adj[node][i]]);
    }
  dep[node] += mx;
}
void dfs2(int node, int par) {
  int k1 = 0;
  int k2 = 0;
  int coun = 0;
  for (int i = 0; i < adj[node].size(); i++)
    if (adj[node][i] != par) {
      if (dep[adj[node][i]] == 1) coun++;
      if (dep[adj[node][i]] % 2 == 0)
        k1 = 1;
      else
        k2 = 1;
    }
  minu += max(0, coun - 1);
  if (k1 == 1 && k2 == 1) f = 3;
  for (int i = 0; i < adj[node].size(); i++)
    if (adj[node][i] != par) dfs2(adj[node][i], node);
}
int main() {
  minu = 0;
  f = 1;
  int n;
  cin >> n;
  int u, v;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int root;
  for (int i = 1; i <= n; i++)
    if (adj[i].size() > 1) {
      root = i;
      break;
    }
  dfs(root, -1);
  dfs2(root, -1);
  cout << f << " " << n - 1 - minu;
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
vector<int> adj[100005];
int dep[100005];
int minu;
int f;
void dfs(int node, int par) {
  dep[node] = 1;
  int mx = 0;
  for (int i = 0; i < adj[node].size(); i++)
    if (adj[node][i] != par) {
      dfs(adj[node][i], node);
      mx = max(mx, dep[adj[node][i]]);
    }
  dep[node] += mx;
}
void dfs2(int node, int par) {
  int k1 = 0;
  int k2 = 0;
  int coun = 0;
  for (int i = 0; i < adj[node].size(); i++)
    if (adj[node][i] != par) {
      if (dep[adj[node][i]] == 1) coun++;
      if (dep[adj[node][i]] % 2 == 0)
        k1 = 1;
      else
        k2 = 1;
    }
  minu += max(0, coun - 1);
  if (k1 == 1 && k2 == 1) f = 3;
  for (int i = 0; i < adj[node].size(); i++)
    if (adj[node][i] != par) dfs2(adj[node][i], node);
}
int main() {
  minu = 0;
  f = 1;
  int n;
  cin >> n;
  int u, v;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int root;
  for (int i = 1; i <= n; i++)
    if (adj[i].size() > 1) {
      root = i;
      break;
    }
  dfs(root, -1);
  dfs2(root, -1);
  cout << f << " " << n - 1 - minu;
  return 0;
}

```