#include <bits/stdc++.h>
using namespace std;
long long n, k;
vector<long long> v;
bool is = false;
vector<vector<long long>> graph;
void dfs(long long i, long long par, long long dis = 0) {
  if (graph[i].size() == 1 && dis % 2 == 1) is = true;
  for (auto j : graph[i]) {
    if (j != par) dfs(j, i, dis + 1);
  }
}
void dfs2(long long i, long long par) {
  long long temp = 0;
  for (auto j : graph[i]) {
    if (j != par) {
      dfs2(j, i);
    }
    if (graph[j].size() == 1) temp += 1;
  }
  if (temp > 0) k += 1;
}
void solve() {
  cin >> n;
  graph.resize(n + 1);
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    graph[y].push_back(x);
    graph[x].push_back(y);
  }
  long long node;
  vector<long long> leafnode;
  for (long long i = 1; i <= n; i++) {
    if (graph[i].size() == 1) {
      node = i;
      break;
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (graph[i].size() == 1) leafnode.push_back(i);
  }
  dfs(node, -1);
  if (is) cout << 3 << " ";
  if (!is) cout << 1 << " ";
  dfs2(node, -1);
  cout << (n - 1) + k - (leafnode.size());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q = 1;
  while (q--) {
    v.clear();
    solve();
  }
}

### 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;
long long n, k;
vector<long long> v;
bool is = false;
vector<vector<long long>> graph;
void dfs(long long i, long long par, long long dis = 0) {
  if (graph[i].size() == 1 && dis % 2 == 1) is = true;
  for (auto j : graph[i]) {
    if (j != par) dfs(j, i, dis + 1);
  }
}
void dfs2(long long i, long long par) {
  long long temp = 0;
  for (auto j : graph[i]) {
    if (j != par) {
      dfs2(j, i);
    }
    if (graph[j].size() == 1) temp += 1;
  }
  if (temp > 0) k += 1;
}
void solve() {
  cin >> n;
  graph.resize(n + 1);
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    graph[y].push_back(x);
    graph[x].push_back(y);
  }
  long long node;
  vector<long long> leafnode;
  for (long long i = 1; i <= n; i++) {
    if (graph[i].size() == 1) {
      node = i;
      break;
    }
  }
  for (long long i = 1; i <= n; i++) {
    if (graph[i].size() == 1) leafnode.push_back(i);
  }
  dfs(node, -1);
  if (is) cout << 3 << " ";
  if (!is) cout << 1 << " ";
  dfs2(node, -1);
  cout << (n - 1) + k - (leafnode.size());
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long q = 1;
  while (q--) {
    v.clear();
    solve();
  }
}

```