#include <bits/stdc++.h>
using namespace std;
vector<int> tree[(int)(1e5 + 5)], h(100005), p(100005);
void dfs(int curr, int parent) {
  h[curr] = h[parent] + 1;
  p[curr] = parent;
  for (int ele : tree[curr]) {
    if (ele == parent) continue;
    dfs(ele, curr);
  }
}
int main() {
  int n, t, f;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> t >> f;
    tree[t].push_back(f);
    tree[f].push_back(t);
  }
  int leaf;
  for (int i = 1; i <= n; i++) {
    if (tree[i].size() == 1) {
      leaf = i;
      break;
    }
  }
  dfs(leaf, 0);
  int o = 0, e = 0;
  map<int, int> mp;
  mp[tree[leaf][0]] = 1;
  for (int i = 1; i <= n; i++) {
    if (tree[i].size() == 1) {
      if (h[i] % 2 == 1)
        e = 1;
      else
        o = 1;
      mp[p[i]]++;
    }
  }
  int a1, a2 = n - 1;
  if (e) a1 = 1;
  if (o) a1 = 3;
  for (int i = 1; i <= n; i++) {
    if (mp[i] != 0) {
      a2 -= (mp[i] - 1);
    }
  }
  cout << a1 << " " << a2;
  return 0;
}

### Prompt

Please formulate 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> tree[(int)(1e5 + 5)], h(100005), p(100005);
void dfs(int curr, int parent) {
  h[curr] = h[parent] + 1;
  p[curr] = parent;
  for (int ele : tree[curr]) {
    if (ele == parent) continue;
    dfs(ele, curr);
  }
}
int main() {
  int n, t, f;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> t >> f;
    tree[t].push_back(f);
    tree[f].push_back(t);
  }
  int leaf;
  for (int i = 1; i <= n; i++) {
    if (tree[i].size() == 1) {
      leaf = i;
      break;
    }
  }
  dfs(leaf, 0);
  int o = 0, e = 0;
  map<int, int> mp;
  mp[tree[leaf][0]] = 1;
  for (int i = 1; i <= n; i++) {
    if (tree[i].size() == 1) {
      if (h[i] % 2 == 1)
        e = 1;
      else
        o = 1;
      mp[p[i]]++;
    }
  }
  int a1, a2 = n - 1;
  if (e) a1 = 1;
  if (o) a1 = 3;
  for (int i = 1; i <= n; i++) {
    if (mp[i] != 0) {
      a2 -= (mp[i] - 1);
    }
  }
  cout << a1 << " " << a2;
  return 0;
}

```