#include <bits/stdc++.h>
using namespace std;
void solve();
void solveA();
void solveB();
void solveC();
void solveD();
void solveE();
void solveF();
void solveG();
void solveTests() {
  int nTests;
  cin >> nTests;
  for (int i = 0; i < nTests; i++) {
    solve();
  }
}
template <typename T>
T get_max_in_vector(vector<T> &v) {
  T mx = v[0];
  for (T tmp : v) mx = max(mx, tmp);
  return mx;
}
template <typename T>
T get_min_in_vector(vector<T> &v) {
  T mn = v[0];
  for (T tmp : v) mn = min(mn, tmp);
  return mn;
}
int main() {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}
void solve() { solveA(); }
void informatics_282() {}
vector<int> g[(int)1e5];
bool been[(int)1e5];
int l[(int)1e5];
int n;
bool leafs_are_even = false;
bool leafs_are_odd = false;
void dfs(int vertex, int level) {
  if (been[vertex]) return;
  been[vertex] = true;
  if (g[vertex].size() == 1) {
    if (level % 2 == 0)
      leafs_are_even = true;
    else
      leafs_are_odd = true;
  }
  for (int i = 0; i < g[vertex].size(); i++) {
    if (!been[g[vertex][i]]) dfs(g[vertex][i], level + 1);
  }
}
int get_num_child_leafs(int vert) {
  int count = 0;
  for (int i = 0; i < g[vert].size(); i++) {
    count += (g[g[vert][i]].size() == 1);
  }
  return count;
}
void solveA() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int mx = n - 1;
  dfs(0, 0);
  for (int i = 0; i < n; i++) {
    int ncl = get_num_child_leafs(i);
    if (ncl > 1) mx -= ncl - 1;
  }
  int mn = (leafs_are_odd && leafs_are_even) ? 3 : 1;
  cout << mn << " " << mx << endl;
}
void solveB() {}
void solveC() {}
void solveD() {}
void solveE() {}

### Prompt

Generate 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;
void solve();
void solveA();
void solveB();
void solveC();
void solveD();
void solveE();
void solveF();
void solveG();
void solveTests() {
  int nTests;
  cin >> nTests;
  for (int i = 0; i < nTests; i++) {
    solve();
  }
}
template <typename T>
T get_max_in_vector(vector<T> &v) {
  T mx = v[0];
  for (T tmp : v) mx = max(mx, tmp);
  return mx;
}
template <typename T>
T get_min_in_vector(vector<T> &v) {
  T mn = v[0];
  for (T tmp : v) mn = min(mn, tmp);
  return mn;
}
int main() {
  ios::sync_with_stdio(false);
  solve();
  return 0;
}
void solve() { solveA(); }
void informatics_282() {}
vector<int> g[(int)1e5];
bool been[(int)1e5];
int l[(int)1e5];
int n;
bool leafs_are_even = false;
bool leafs_are_odd = false;
void dfs(int vertex, int level) {
  if (been[vertex]) return;
  been[vertex] = true;
  if (g[vertex].size() == 1) {
    if (level % 2 == 0)
      leafs_are_even = true;
    else
      leafs_are_odd = true;
  }
  for (int i = 0; i < g[vertex].size(); i++) {
    if (!been[g[vertex][i]]) dfs(g[vertex][i], level + 1);
  }
}
int get_num_child_leafs(int vert) {
  int count = 0;
  for (int i = 0; i < g[vert].size(); i++) {
    count += (g[g[vert][i]].size() == 1);
  }
  return count;
}
void solveA() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int mx = n - 1;
  dfs(0, 0);
  for (int i = 0; i < n; i++) {
    int ncl = get_num_child_leafs(i);
    if (ncl > 1) mx -= ncl - 1;
  }
  int mn = (leafs_are_odd && leafs_are_even) ? 3 : 1;
  cout << mn << " " << mx << endl;
}
void solveB() {}
void solveC() {}
void solveD() {}
void solveE() {}

```