#include <bits/stdc++.h>
using namespace std;
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &item) {
  out << '(' << item.first << ", " << item.second << ')';
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (const auto &item : v) out << item << ' ';
  return out;
}
using namespace std;
int n;
vector<int> G[100010];
int vis[100010], f[100010];
int odd, even;
void dfs(int node, int depth = 0) {
  vis[node] = 1;
  bool leaf = true;
  for (auto next : G[node]) {
    if (!vis[next]) {
      leaf = false;
      dfs(next, depth + 1);
    }
  }
  if (leaf) {
    if (depth % 2)
      odd++;
    else
      even++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int root = -1;
  int leafs = 0;
  int parOfLeafs = 0;
  for (int i = 1; i <= n; i++) {
    if (G[i].size() > 1) root = i;
    if (G[i].size() == 1) {
      leafs++;
      if (!f[G[i][0]]) f[G[i][0]] = 1, parOfLeafs++;
    }
  }
  dfs(root);
  if (odd != 0 && even != 0)
    cout << 3 << ' ';
  else
    cout << 1 << ' ';
  cout << (n - 1) - leafs + parOfLeafs;
  return 0;
}

### Prompt

Your task is to create 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;
template <typename T1, typename T2>
ostream &operator<<(ostream &out, const pair<T1, T2> &item) {
  out << '(' << item.first << ", " << item.second << ')';
  return out;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  for (const auto &item : v) out << item << ' ';
  return out;
}
using namespace std;
int n;
vector<int> G[100010];
int vis[100010], f[100010];
int odd, even;
void dfs(int node, int depth = 0) {
  vis[node] = 1;
  bool leaf = true;
  for (auto next : G[node]) {
    if (!vis[next]) {
      leaf = false;
      dfs(next, depth + 1);
    }
  }
  if (leaf) {
    if (depth % 2)
      odd++;
    else
      even++;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int root = -1;
  int leafs = 0;
  int parOfLeafs = 0;
  for (int i = 1; i <= n; i++) {
    if (G[i].size() > 1) root = i;
    if (G[i].size() == 1) {
      leafs++;
      if (!f[G[i][0]]) f[G[i][0]] = 1, parOfLeafs++;
    }
  }
  dfs(root);
  if (odd != 0 && even != 0)
    cout << 3 << ' ';
  else
    cout << 1 << ' ';
  cout << (n - 1) - leafs + parOfLeafs;
  return 0;
}

```