#include <bits/stdc++.h>
inline int read() {
  char c = getchar();
  int x = 0, ft = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') ft = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  return x * ft;
}
const int N = 1e5 + 5;
int n, rt, flag;
std::vector<int> e[N];
inline void add_edge(int u, int v) {
  e[u].push_back(v);
  e[v].push_back(u);
}
void dfs(int u, int pre, int dep) {
  if ((int)e[u].size() == 1 && dep % 2 == 1) flag = 1;
  for (int i = 0; i < (int)e[u].size(); ++i) {
    int v = e[u][i];
    if (v != pre) dfs(v, u, dep + 1);
  }
}
int main() {
  n = read();
  for (int i = 1; i < n; ++i) add_edge(read(), read());
  for (int i = 1; i <= n; ++i)
    if ((int)e[i].size() == 1) {
      rt = i;
      break;
    }
  dfs(rt, 0, 0);
  printf("%d ", flag ? 3 : 1);
  int ans = n - 1;
  for (int u = 1; u <= n; ++u) {
    int res = 0;
    for (int i = 0; i < (int)e[u].size(); ++i) {
      int v = e[u][i];
      if ((int)e[v].size() == 1) ++res;
    }
    ans -= res;
    if (res) ++ans;
  }
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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>
inline int read() {
  char c = getchar();
  int x = 0, ft = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') ft = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  return x * ft;
}
const int N = 1e5 + 5;
int n, rt, flag;
std::vector<int> e[N];
inline void add_edge(int u, int v) {
  e[u].push_back(v);
  e[v].push_back(u);
}
void dfs(int u, int pre, int dep) {
  if ((int)e[u].size() == 1 && dep % 2 == 1) flag = 1;
  for (int i = 0; i < (int)e[u].size(); ++i) {
    int v = e[u][i];
    if (v != pre) dfs(v, u, dep + 1);
  }
}
int main() {
  n = read();
  for (int i = 1; i < n; ++i) add_edge(read(), read());
  for (int i = 1; i <= n; ++i)
    if ((int)e[i].size() == 1) {
      rt = i;
      break;
    }
  dfs(rt, 0, 0);
  printf("%d ", flag ? 3 : 1);
  int ans = n - 1;
  for (int u = 1; u <= n; ++u) {
    int res = 0;
    for (int i = 0; i < (int)e[u].size(); ++i) {
      int v = e[u][i];
      if ((int)e[v].size() == 1) ++res;
    }
    ans -= res;
    if (res) ++ans;
  }
  printf("%d\n", ans);
  return 0;
}

```