#include <bits/stdc++.h>
using namespace std;
vector<long long int> v[1000000];
long long int mark[1000000], tt = 1;
long long int find(long long int node) {
  if (v[node].size() == 1 && node != 1) return 1;
  mark[node] = 1;
  long long int mk1 = 0, mk2 = 0;
  for (long long int i = 0; i < v[node].size(); i++) {
    if (!mark[v[node][i]]) {
      long long int x = find(v[node][i]);
      if (x == 1)
        mk1++;
      else
        mk2++;
    }
  }
  if (node == 1) {
    if (v[node].size() == 1) {
      if (mk1) tt = 3;
      return 1;
    }
  }
  if (mk2 != 0 && mk1 != 0) {
    tt = 3;
    return 0;
  }
  if (mk2 == 0)
    return 0;
  else
    return 1;
}
long long int dfs(long long int node) {
  if (v[node].size() == 1 && node != 1) return 1;
  mark[node] = 1;
  long long int mk = 0, count = 0;
  for (long long int i = 0; i < v[node].size(); i++) {
    if (!mark[v[node][i]]) {
      long long int x = dfs(v[node][i]);
      if (x != 1)
        count += x;
      else
        mk = 1;
    }
  }
  return count + mk + 1;
}
int main() {
  long long int n, x, y;
  cin >> n;
  for (long long int i = 1; i < n; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  long long int p = find(1);
  for (long long int i = 0; i <= n; i++) mark[i] = 0;
  cout << tt << " ";
  long long int count = 0;
  if (v[1].size() == 1) {
    x = v[1][0];
    for (long long int j = 0; j < v[x].size(); j++) {
      y = v[x][j];
      if (v[y].size() == 1 && y != 1) {
        count = 1;
      }
    }
  }
  cout << dfs(1) - 1 - count;
}

### 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;
vector<long long int> v[1000000];
long long int mark[1000000], tt = 1;
long long int find(long long int node) {
  if (v[node].size() == 1 && node != 1) return 1;
  mark[node] = 1;
  long long int mk1 = 0, mk2 = 0;
  for (long long int i = 0; i < v[node].size(); i++) {
    if (!mark[v[node][i]]) {
      long long int x = find(v[node][i]);
      if (x == 1)
        mk1++;
      else
        mk2++;
    }
  }
  if (node == 1) {
    if (v[node].size() == 1) {
      if (mk1) tt = 3;
      return 1;
    }
  }
  if (mk2 != 0 && mk1 != 0) {
    tt = 3;
    return 0;
  }
  if (mk2 == 0)
    return 0;
  else
    return 1;
}
long long int dfs(long long int node) {
  if (v[node].size() == 1 && node != 1) return 1;
  mark[node] = 1;
  long long int mk = 0, count = 0;
  for (long long int i = 0; i < v[node].size(); i++) {
    if (!mark[v[node][i]]) {
      long long int x = dfs(v[node][i]);
      if (x != 1)
        count += x;
      else
        mk = 1;
    }
  }
  return count + mk + 1;
}
int main() {
  long long int n, x, y;
  cin >> n;
  for (long long int i = 1; i < n; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  long long int p = find(1);
  for (long long int i = 0; i <= n; i++) mark[i] = 0;
  cout << tt << " ";
  long long int count = 0;
  if (v[1].size() == 1) {
    x = v[1][0];
    for (long long int j = 0; j < v[x].size(); j++) {
      y = v[x][j];
      if (v[y].size() == 1 && y != 1) {
        count = 1;
      }
    }
  }
  cout << dfs(1) - 1 - count;
}

```