#include <bits/stdc++.h>
using namespace std;
int degree[100000 + 1] = {0};
void dfs1(int u, vector<int> edges[], int p, int &maxim, int traced = 0) {
  bool attached_leaf = false;
  for (auto v : edges[u]) {
    if (v != p) {
      if (degree[v] == 1) {
        if (traced > 1) {
          attached_leaf = true;
        }
        continue;
      }
      maxim++;
      traced++;
      dfs1(v, edges, u, maxim, traced);
      traced--;
    }
  }
  if (attached_leaf) {
    ++maxim;
  }
  return;
}
bool dfs2(int start, vector<int> edges[], int p, int count) {
  if (p != -1 && degree[start] == 1) {
    if (count & 1) {
      return true;
    }
    return false;
  }
  bool ans = false;
  for (auto v : edges[start]) {
    if (v != p) {
      ans = ans || dfs2(v, edges, start, count + 1);
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> edges[n + 1];
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    edges[u].push_back(v);
    edges[v].push_back(u);
    degree[u]++;
    degree[v]++;
  }
  int start = 1;
  for (int i = 1; i <= n; i++) {
    if (degree[i] == 1) {
      start = i;
      break;
    }
  }
  int maxim = 0, minim = 1;
  dfs1(start, edges, -1, maxim, 0);
  bool ans = dfs2(start, edges, -1, 0);
  if (ans) {
    minim = 3;
  }
  cout << minim << " " << maxim << 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 degree[100000 + 1] = {0};
void dfs1(int u, vector<int> edges[], int p, int &maxim, int traced = 0) {
  bool attached_leaf = false;
  for (auto v : edges[u]) {
    if (v != p) {
      if (degree[v] == 1) {
        if (traced > 1) {
          attached_leaf = true;
        }
        continue;
      }
      maxim++;
      traced++;
      dfs1(v, edges, u, maxim, traced);
      traced--;
    }
  }
  if (attached_leaf) {
    ++maxim;
  }
  return;
}
bool dfs2(int start, vector<int> edges[], int p, int count) {
  if (p != -1 && degree[start] == 1) {
    if (count & 1) {
      return true;
    }
    return false;
  }
  bool ans = false;
  for (auto v : edges[start]) {
    if (v != p) {
      ans = ans || dfs2(v, edges, start, count + 1);
    }
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  vector<int> edges[n + 1];
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    edges[u].push_back(v);
    edges[v].push_back(u);
    degree[u]++;
    degree[v]++;
  }
  int start = 1;
  for (int i = 1; i <= n; i++) {
    if (degree[i] == 1) {
      start = i;
      break;
    }
  }
  int maxim = 0, minim = 1;
  dfs1(start, edges, -1, maxim, 0);
  bool ans = dfs2(start, edges, -1, 0);
  if (ans) {
    minim = 3;
  }
  cout << minim << " " << maxim << endl;
}

```