#include <bits/stdc++.h>
using namespace std;
int const N = 1e5 + 1;
vector<int> v[N];
int maxi = 0;
int dp[N];
int mini = 1;
void dfs(int u, int p) {
  int now = 0;
  bool leaf_child = false;
  bool odd = false;
  bool even = false;
  for (int a : v[u]) {
    if (a == p) {
      continue;
    }
    dfs(a, u);
    now = max(now, dp[a]);
    if (leaf_child == true && dp[a] == 1) {
      maxi--;
    }
    if (dp[a] == 1) {
      leaf_child = true;
    }
    if (dp[a] % 2 == 1) {
      odd = true;
    }
    if (dp[a] % 2 == 0) {
      even = true;
    }
  }
  if (odd && even) {
    mini = 3;
  }
  dp[u] = now + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    int ind;
    for (int i = 1; i <= n; i++) {
      if (v[i].size() > 1) {
        ind = i;
        break;
      }
    }
    dfs(ind, -1);
    cout << mini << " " << maxi + n - 1 << endl;
  }
}

### 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;
int const N = 1e5 + 1;
vector<int> v[N];
int maxi = 0;
int dp[N];
int mini = 1;
void dfs(int u, int p) {
  int now = 0;
  bool leaf_child = false;
  bool odd = false;
  bool even = false;
  for (int a : v[u]) {
    if (a == p) {
      continue;
    }
    dfs(a, u);
    now = max(now, dp[a]);
    if (leaf_child == true && dp[a] == 1) {
      maxi--;
    }
    if (dp[a] == 1) {
      leaf_child = true;
    }
    if (dp[a] % 2 == 1) {
      odd = true;
    }
    if (dp[a] % 2 == 0) {
      even = true;
    }
  }
  if (odd && even) {
    mini = 3;
  }
  dp[u] = now + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    int ind;
    for (int i = 1; i <= n; i++) {
      if (v[i].size() > 1) {
        ind = i;
        break;
      }
    }
    dfs(ind, -1);
    cout << mini << " " << maxi + n - 1 << endl;
  }
}

```