#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const long long mn = 2e5;
long long n, a, b, mini = 1, maxi, root, k1[mn], k2[mn];
vector<long long> v[mn];
void dfs(long long i, long long pr = -1) {
  long long cnt = 0;
  for (long long j : v[i])
    if (j != pr) {
      if (v[j].size() == 1)
        cnt++;
      else {
        dfs(j, i);
        k1[i] += k2[j];
        k2[i] += k1[j];
      }
    }
  k1[i] += cnt;
  if (k1[i] > 0 && k2[i] > 0) mini = 3;
  maxi -= max(0ll, cnt - 1);
}
signed main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i < n; i++) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    if (v[i].size() > 1) root = i;
  }
  maxi = n - 1;
  dfs(root);
  cout << mini << " " << maxi;
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("O3")
using namespace std;
const long long mn = 2e5;
long long n, a, b, mini = 1, maxi, root, k1[mn], k2[mn];
vector<long long> v[mn];
void dfs(long long i, long long pr = -1) {
  long long cnt = 0;
  for (long long j : v[i])
    if (j != pr) {
      if (v[j].size() == 1)
        cnt++;
      else {
        dfs(j, i);
        k1[i] += k2[j];
        k2[i] += k1[j];
      }
    }
  k1[i] += cnt;
  if (k1[i] > 0 && k2[i] > 0) mini = 3;
  maxi -= max(0ll, cnt - 1);
}
signed main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (long long i = 1; i < n; i++) {
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    if (v[i].size() > 1) root = i;
  }
  maxi = n - 1;
  dfs(root);
  cout << mini << " " << maxi;
  return 0;
}

```