#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
int n;
vector<int> nei[N + 1];
int dep[N + 1];
vector<int> deps;
int root;
set<int> st;
vector<int> vv;
void dfs(int x = root, int fa = 0) {
  if (nei[x].size() == 1 && x != root)
    deps.push_back(dep[x]), dep[x] != 2 && (st.insert(fa), vv.push_back(fa), 0);
  for (int i = 0; i < nei[x].size(); i++) {
    int y = nei[x][i];
    if (y == fa) continue;
    dep[y] = dep[x] + 1;
    dfs(y, x);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    nei[x].push_back(y);
    nei[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (nei[i].size() == 1) root = i;
  dfs();
  int ans1 = 1, ans2 = n - 1 - (vv.size() - st.size());
  for (int i = 0; i < deps.size(); i++)
    if (deps[i] & 1) ans1 = 3;
  for (int i = 0; i < deps.size(); i++) ans2 -= deps[i] == 2;
  cout << ans1 << " " << ans2;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
const int N = 100000;
int n;
vector<int> nei[N + 1];
int dep[N + 1];
vector<int> deps;
int root;
set<int> st;
vector<int> vv;
void dfs(int x = root, int fa = 0) {
  if (nei[x].size() == 1 && x != root)
    deps.push_back(dep[x]), dep[x] != 2 && (st.insert(fa), vv.push_back(fa), 0);
  for (int i = 0; i < nei[x].size(); i++) {
    int y = nei[x][i];
    if (y == fa) continue;
    dep[y] = dep[x] + 1;
    dfs(y, x);
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    nei[x].push_back(y);
    nei[y].push_back(x);
  }
  for (int i = 1; i <= n; i++)
    if (nei[i].size() == 1) root = i;
  dfs();
  int ans1 = 1, ans2 = n - 1 - (vv.size() - st.size());
  for (int i = 0; i < deps.size(); i++)
    if (deps[i] & 1) ans1 = 3;
  for (int i = 0; i < deps.size(); i++) ans2 -= deps[i] == 2;
  cout << ans1 << " " << ans2;
  return 0;
}

```