#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b,
                    long long int M = 998244353) {
  long long int result = 1;
  for (;;) {
    if (b & 1) result = (result * a) % M;
    b >>= 1;
    if (!b) break;
    a = (a * a) % M;
  }
  return result;
}
int minn = 0;
void dfs(vector<vector<long long int> >& graph, int mark[], long long int root,
         long long int par[], long long int h) {
  int flag = 0;
  mark[root] = 1;
  for (auto i : graph[root])
    if (!mark[i]) par[i] = root, dfs(graph, mark, i, par, h + 1), flag = 1;
  if (flag == 0) minn |= h % 2 + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  vector<vector<long long int> > graph(n + 1);
  long long int d[n + 1];
  memset(d, 0, sizeof(d));
  for (long long int i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
    d[x] += 1, d[y] += 1;
  }
  long long int root = -1;
  for (long long int i = 1; i < n + 1; i++) {
    if (d[i] > 1) root = i;
  }
  if (root == -1) {
    cout << 1 << " " << 1 << endl;
    return 0;
  }
  long long int one = 1, two = n - 1;
  int mark[n + 1];
  memset(mark, 0, sizeof(mark));
  long long int par[n + 1];
  memset(par, 0, sizeof(par));
  dfs(graph, mark, root, par, 0);
  long long int cnt[n + 1];
  memset(cnt, 0, sizeof(cnt));
  for (long long int i = 1; i < n + 1; i++) {
    if (d[i] == 1) cnt[par[i]]++;
  }
  for (long long int i = 1; i < n + 1; i++)
    if (cnt[i] != 0) two -= (cnt[i] - 1);
  one = minn - ((minn == 2) ? 1 : 0);
  cout << one << " " << two << endl;
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
long long int power(long long int a, long long int b,
                    long long int M = 998244353) {
  long long int result = 1;
  for (;;) {
    if (b & 1) result = (result * a) % M;
    b >>= 1;
    if (!b) break;
    a = (a * a) % M;
  }
  return result;
}
int minn = 0;
void dfs(vector<vector<long long int> >& graph, int mark[], long long int root,
         long long int par[], long long int h) {
  int flag = 0;
  mark[root] = 1;
  for (auto i : graph[root])
    if (!mark[i]) par[i] = root, dfs(graph, mark, i, par, h + 1), flag = 1;
  if (flag == 0) minn |= h % 2 + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  vector<vector<long long int> > graph(n + 1);
  long long int d[n + 1];
  memset(d, 0, sizeof(d));
  for (long long int i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
    d[x] += 1, d[y] += 1;
  }
  long long int root = -1;
  for (long long int i = 1; i < n + 1; i++) {
    if (d[i] > 1) root = i;
  }
  if (root == -1) {
    cout << 1 << " " << 1 << endl;
    return 0;
  }
  long long int one = 1, two = n - 1;
  int mark[n + 1];
  memset(mark, 0, sizeof(mark));
  long long int par[n + 1];
  memset(par, 0, sizeof(par));
  dfs(graph, mark, root, par, 0);
  long long int cnt[n + 1];
  memset(cnt, 0, sizeof(cnt));
  for (long long int i = 1; i < n + 1; i++) {
    if (d[i] == 1) cnt[par[i]]++;
  }
  for (long long int i = 1; i < n + 1; i++)
    if (cnt[i] != 0) two -= (cnt[i] - 1);
  one = minn - ((minn == 2) ? 1 : 0);
  cout << one << " " << two << endl;
}

```