#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> adj[n];
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int rt = 0;
  for (int i = 0; i < n; i++) {
    if (adj[n].size() > 1) {
      rt = i;
    }
  }
  int ds[n];
  bool vis[n];
  for (int i = 0; i < n; i++) {
    ds[i] = 1000000000;
  }
  memset(vis, false, sizeof vis);
  queue<pair<int, int> > inf;
  inf.push({rt, 0});
  while (inf.size()) {
    pair<int, int> info = inf.front();
    inf.pop();
    int vt = info.first;
    if (vis[vt]) {
      continue;
    }
    vis[vt] = true;
    int dst = info.second;
    ds[vt] = dst;
    for (int a : adj[vt]) {
      if (!vis[a]) {
        inf.push({a, dst + 1});
      }
    }
  }
  bool f = false;
  bool g = false;
  int l = 0;
  int m = 0;
  for (int i = 0; i < n; i++) {
    bool h = false;
    if (adj[i].size() == 1) {
      l++;
      if (ds[i] % 2 == 0) {
        f = true;
      } else {
        g = true;
      }
    } else {
      for (int j = 0; j < adj[i].size(); j++) {
        if (adj[adj[i][j]].size() == 1) {
          h = true;
        }
      }
    }
    if (h) {
      m++;
    }
  }
  int mn = 1000000000;
  if ((f && !g) || (g && !f)) {
    mn = 1;
  } else {
    mn = 3;
  }
  int mx = n - 1 - l + m;
  cout << mn << " " << mx << endl;
}
void querySolve() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}

### Prompt

In Cpp, your task is to solve 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;
void solve() {
  int n;
  cin >> n;
  vector<int> adj[n];
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  int rt = 0;
  for (int i = 0; i < n; i++) {
    if (adj[n].size() > 1) {
      rt = i;
    }
  }
  int ds[n];
  bool vis[n];
  for (int i = 0; i < n; i++) {
    ds[i] = 1000000000;
  }
  memset(vis, false, sizeof vis);
  queue<pair<int, int> > inf;
  inf.push({rt, 0});
  while (inf.size()) {
    pair<int, int> info = inf.front();
    inf.pop();
    int vt = info.first;
    if (vis[vt]) {
      continue;
    }
    vis[vt] = true;
    int dst = info.second;
    ds[vt] = dst;
    for (int a : adj[vt]) {
      if (!vis[a]) {
        inf.push({a, dst + 1});
      }
    }
  }
  bool f = false;
  bool g = false;
  int l = 0;
  int m = 0;
  for (int i = 0; i < n; i++) {
    bool h = false;
    if (adj[i].size() == 1) {
      l++;
      if (ds[i] % 2 == 0) {
        f = true;
      } else {
        g = true;
      }
    } else {
      for (int j = 0; j < adj[i].size(); j++) {
        if (adj[adj[i][j]].size() == 1) {
          h = true;
        }
      }
    }
    if (h) {
      m++;
    }
  }
  int mn = 1000000000;
  if ((f && !g) || (g && !f)) {
    mn = 1;
  } else {
    mn = 3;
  }
  int mx = n - 1 - l + m;
  cout << mn << " " << mx << endl;
}
void querySolve() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  solve();
}

```