#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> v(n);
  for (int i = 0; i < (int)(n - 1); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  vector<int> dp(n, 1e9);
  dp[0] = 0;
  queue<int> que;
  que.push(0);
  while (!que.empty()) {
    int id = que.front();
    que.pop();
    int d = dp[id];
    for (auto i : v[id]) {
      if (dp[i] > d + 1) {
        dp[i] = d + 1;
        que.push(i);
      }
    }
  }
  set<int> res_mn;
  for (int i = 0; i < n; i++) {
    if (v[i].size() == 1) {
      res_mn.insert(dp[i] % 2);
    }
  }
  if (res_mn.size() == 1) {
    cout << 1 << ' ';
  } else {
    if (n == 2) {
      cout << 1 << ' ';
    } else {
      cout << 3 << ' ';
    }
  }
  int mx = n - 1;
  vector<bool> is_leaf(n, false);
  for (int i = 0; i < n; i++) {
    if (v[i].size() == 1) {
      is_leaf[i] = true;
    }
  }
  for (int i = 0; i < n; i++) {
    int cnt = 0;
    for (auto j : v[i]) {
      cnt += is_leaf[j];
    }
    if (cnt > 1) {
      mx -= cnt - 1;
    }
  }
  cout << mx << '\n';
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<vector<int>> v(n);
  for (int i = 0; i < (int)(n - 1); i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  vector<int> dp(n, 1e9);
  dp[0] = 0;
  queue<int> que;
  que.push(0);
  while (!que.empty()) {
    int id = que.front();
    que.pop();
    int d = dp[id];
    for (auto i : v[id]) {
      if (dp[i] > d + 1) {
        dp[i] = d + 1;
        que.push(i);
      }
    }
  }
  set<int> res_mn;
  for (int i = 0; i < n; i++) {
    if (v[i].size() == 1) {
      res_mn.insert(dp[i] % 2);
    }
  }
  if (res_mn.size() == 1) {
    cout << 1 << ' ';
  } else {
    if (n == 2) {
      cout << 1 << ' ';
    } else {
      cout << 3 << ' ';
    }
  }
  int mx = n - 1;
  vector<bool> is_leaf(n, false);
  for (int i = 0; i < n; i++) {
    if (v[i].size() == 1) {
      is_leaf[i] = true;
    }
  }
  for (int i = 0; i < n; i++) {
    int cnt = 0;
    for (auto j : v[i]) {
      cnt += is_leaf[j];
    }
    if (cnt > 1) {
      mx -= cnt - 1;
    }
  }
  cout << mx << '\n';
  return 0;
}

```