#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const int maxn = 1e5 + 9;
const double pi = 3.14159265358979323846264;
const int inf = 2e9;
const long long mod = 1e9 + 9;
vector<int> g[maxn];
int ans;
struct node {
  int fg, len, jk;
  node() {}
  node(int fg, int len, int jk) : fg(fg), len(len), jk(jk) {}
};
node sol(int u, int par) {
  int cnum = g[u].size();
  if (par != -1) --cnum;
  node io;
  if (cnum == 0)
    return node(0, 0, 0);
  else if (cnum == 1) {
    for (int i = 0; i < g[u].size(); ++i) {
      int v = g[u][i];
      if (v != par) {
        node tmp = sol(v, u);
        if (!tmp.fg) {
          io.fg = 0;
          io.len = tmp.len + 1;
        } else
          io.fg = 1, ++ans;
        if (!tmp.jk)
          io.jk = 1;
        else if (tmp.jk == 1)
          io.jk = 0;
        else
          io.jk = 2;
      }
    }
  } else {
    io.fg = 1;
    io.jk = -1;
    io.len = 0;
    int ui = 0;
    for (int i = 0; i < g[u].size(); ++i) {
      int v = g[u][i];
      if (v != par) {
        node tmp = sol(v, u);
        if (!tmp.fg) {
          int kl = tmp.len + 1;
          if (kl > 1)
            ans += kl;
          else
            ++ui;
        } else
          ++ans;
        if (io.jk < 0) {
          if (tmp.jk == 0)
            io.jk = 1;
          else if (tmp.jk == 1)
            io.jk = 0;
          else
            io.jk = 2;
        } else if (io.jk < 2) {
          if (tmp.jk == io.jk || tmp.jk == 2) io.jk = 2;
        }
      }
    }
    if (ui) ++ans;
  }
  return io;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int root;
  for (int i = 1; i <= n; ++i) {
    if (g[i].size() > 1) {
      root = i;
      break;
    }
  }
  ans = 0;
  node tmp = sol(root, -1);
  if (tmp.jk < 2)
    cout << "1 ";
  else
    cout << "3 ";
  cout << ans;
  return 0;
}

### Prompt

Please provide a cpp coded solution 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;
const double eps = 1e-9;
const int maxn = 1e5 + 9;
const double pi = 3.14159265358979323846264;
const int inf = 2e9;
const long long mod = 1e9 + 9;
vector<int> g[maxn];
int ans;
struct node {
  int fg, len, jk;
  node() {}
  node(int fg, int len, int jk) : fg(fg), len(len), jk(jk) {}
};
node sol(int u, int par) {
  int cnum = g[u].size();
  if (par != -1) --cnum;
  node io;
  if (cnum == 0)
    return node(0, 0, 0);
  else if (cnum == 1) {
    for (int i = 0; i < g[u].size(); ++i) {
      int v = g[u][i];
      if (v != par) {
        node tmp = sol(v, u);
        if (!tmp.fg) {
          io.fg = 0;
          io.len = tmp.len + 1;
        } else
          io.fg = 1, ++ans;
        if (!tmp.jk)
          io.jk = 1;
        else if (tmp.jk == 1)
          io.jk = 0;
        else
          io.jk = 2;
      }
    }
  } else {
    io.fg = 1;
    io.jk = -1;
    io.len = 0;
    int ui = 0;
    for (int i = 0; i < g[u].size(); ++i) {
      int v = g[u][i];
      if (v != par) {
        node tmp = sol(v, u);
        if (!tmp.fg) {
          int kl = tmp.len + 1;
          if (kl > 1)
            ans += kl;
          else
            ++ui;
        } else
          ++ans;
        if (io.jk < 0) {
          if (tmp.jk == 0)
            io.jk = 1;
          else if (tmp.jk == 1)
            io.jk = 0;
          else
            io.jk = 2;
        } else if (io.jk < 2) {
          if (tmp.jk == io.jk || tmp.jk == 2) io.jk = 2;
        }
      }
    }
    if (ui) ++ans;
  }
  return io;
}
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int root;
  for (int i = 1; i <= n; ++i) {
    if (g[i].size() > 1) {
      root = i;
      break;
    }
  }
  ans = 0;
  node tmp = sol(root, -1);
  if (tmp.jk < 2)
    cout << "1 ";
  else
    cout << "3 ";
  cout << ans;
  return 0;
}

```