#include <bits/stdc++.h>
using namespace std;
template <typename T>
T *arr(int n) {
  return (T *)malloc(n * sizeof(T));
}
template <typename T>
T *arr(int n, T val) {
  auto res = arr<T>(n);
  fill(res, res + n, val);
  return res;
}
struct Graph {
  int *head, *eto, *nxt, ptr = 0;
  Graph(int n, int m) {
    head = arr<int>(n, -1);
    eto = arr<int>(m);
    nxt = arr<int>(m);
  }
  void add_edge(int from, int to) {
    eto[ptr++] = to;
    nxt[ptr - 1] = head[from];
    head[from] = ptr - 1;
  }
};
void run();
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) {
    run();
  }
}
int a[2];
void dfs(int v, vector<vector<int>> &gr, int dep = 0, int p = -1) {
  if (gr[v].size() == 1) a[dep % 2]++;
  for (int u : gr[v]) {
    if (u == p) continue;
    dfs(u, gr, dep + 1, v);
  }
}
void run() {
  int n;
  cin >> n;
  vector<vector<int>> gr(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  int mx = 0;
  for (int v = 1; v <= n; v++) {
    if (gr[v].size() == 1) continue;
    int cnt = 0;
    for (int u : gr[v]) {
      cnt += gr[u].size() == 1;
    }
    mx += 1 + (cnt > 0);
  }
  mx--;
  dfs(1, gr);
  int mn = 1;
  if (a[0] && a[1]) mn = 3;
  cout << mn << ' ' << mx << '\n';
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 T>
T *arr(int n) {
  return (T *)malloc(n * sizeof(T));
}
template <typename T>
T *arr(int n, T val) {
  auto res = arr<T>(n);
  fill(res, res + n, val);
  return res;
}
struct Graph {
  int *head, *eto, *nxt, ptr = 0;
  Graph(int n, int m) {
    head = arr<int>(n, -1);
    eto = arr<int>(m);
    nxt = arr<int>(m);
  }
  void add_edge(int from, int to) {
    eto[ptr++] = to;
    nxt[ptr - 1] = head[from];
    head[from] = ptr - 1;
  }
};
void run();
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  int t = 1;
  for (int i = 0; i < t; i++) {
    run();
  }
}
int a[2];
void dfs(int v, vector<vector<int>> &gr, int dep = 0, int p = -1) {
  if (gr[v].size() == 1) a[dep % 2]++;
  for (int u : gr[v]) {
    if (u == p) continue;
    dfs(u, gr, dep + 1, v);
  }
}
void run() {
  int n;
  cin >> n;
  vector<vector<int>> gr(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    gr[a].push_back(b);
    gr[b].push_back(a);
  }
  int mx = 0;
  for (int v = 1; v <= n; v++) {
    if (gr[v].size() == 1) continue;
    int cnt = 0;
    for (int u : gr[v]) {
      cnt += gr[u].size() == 1;
    }
    mx += 1 + (cnt > 0);
  }
  mx--;
  dfs(1, gr);
  int mn = 1;
  if (a[0] && a[1]) mn = 3;
  cout << mn << ' ' << mx << '\n';
}

```