#include <bits/stdc++.h>
using namespace std;
const int N = 100005;
vector<int> v[N];
int ans = 0;
set<int> parity;
int dfs(int node, int par, int level) {
  int leaf = 0;
  int cnt = 0;
  for (auto pk : v[node]) {
    if (pk == par) continue;
    int x = dfs(pk, node, level + 1);
    leaf = 1;
    if (x == 1) cnt++;
  }
  if (cnt) ans = ans + (cnt - 1);
  if (leaf == 0) {
    parity.insert(level % 2);
    return 1;
  }
  return 0;
}
int main() {
  int i, j, k, l, m, n, p;
  cin >> n;
  m = n - 1;
  while (m--) {
    cin >> i >> j;
    v[i].push_back(j);
    v[j].push_back(i);
  }
  int root;
  for (i = 1; i <= n; i++) {
    if (v[i].size() > 1) root = i;
  }
  dfs(root, 0, 0);
  ans = n - 1 - ans;
  if (parity.size() % 2)
    cout << 1 << " ";
  else
    cout << 3 << " ";
  cout << ans << 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;
const int N = 100005;
vector<int> v[N];
int ans = 0;
set<int> parity;
int dfs(int node, int par, int level) {
  int leaf = 0;
  int cnt = 0;
  for (auto pk : v[node]) {
    if (pk == par) continue;
    int x = dfs(pk, node, level + 1);
    leaf = 1;
    if (x == 1) cnt++;
  }
  if (cnt) ans = ans + (cnt - 1);
  if (leaf == 0) {
    parity.insert(level % 2);
    return 1;
  }
  return 0;
}
int main() {
  int i, j, k, l, m, n, p;
  cin >> n;
  m = n - 1;
  while (m--) {
    cin >> i >> j;
    v[i].push_back(j);
    v[j].push_back(i);
  }
  int root;
  for (i = 1; i <= n; i++) {
    if (v[i].size() > 1) root = i;
  }
  dfs(root, 0, 0);
  ans = n - 1 - ans;
  if (parity.size() % 2)
    cout << 1 << " ";
  else
    cout << 3 << " ";
  cout << ans << endl;
  return 0;
}

```