#include <bits/stdc++.h>
using namespace std;
const long long int N = 1e5 + 5;
vector<long long int> v[N];
long long int mx = 0, odd = 0, even = 0, tot = 0;
set<long long int> leaves;
long long int leave[100001] = {0};
void dfs(long long int x, long long int par, long long int level) {
  level++;
  if (v[x].size() == 1) {
    if (level % 2 == 0)
      even++;
    else
      odd++;
  }
  for (auto j : v[x]) {
    if (j == par) continue;
    dfs(j, x, level);
  }
}
void dfs2(long long int x, long long int par) {
  long long int ans = 0;
  for (auto j : v[x]) {
    if (j == par) continue;
    if (leaves.find(j) != leaves.end()) {
      ans++;
    }
    dfs2(j, x);
  }
  leave[x] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int i, x, y;
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (i = 1; i <= n; i++) {
    if (v[i].size() == 1) {
      leaves.insert(i);
      tot++;
    }
  }
  dfs(1, 1, 0);
  dfs2(1, -1);
  long long int temp = 0;
  if (leaves.find(1) != leaves.end()) {
    if (v[1].size()) leave[v[1][0]]++;
  }
  for (long long int i = 1; i <= n; i++) {
    if (leave[i]) temp += leave[i] - 1;
  }
  if (odd == tot || even == tot)
    cout << 1 << " ";
  else
    cout << 3 << " ";
  cout << n - 1 - temp << " ";
  return 0;
}

### 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;
const long long int N = 1e5 + 5;
vector<long long int> v[N];
long long int mx = 0, odd = 0, even = 0, tot = 0;
set<long long int> leaves;
long long int leave[100001] = {0};
void dfs(long long int x, long long int par, long long int level) {
  level++;
  if (v[x].size() == 1) {
    if (level % 2 == 0)
      even++;
    else
      odd++;
  }
  for (auto j : v[x]) {
    if (j == par) continue;
    dfs(j, x, level);
  }
}
void dfs2(long long int x, long long int par) {
  long long int ans = 0;
  for (auto j : v[x]) {
    if (j == par) continue;
    if (leaves.find(j) != leaves.end()) {
      ans++;
    }
    dfs2(j, x);
  }
  leave[x] = ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  long long int i, x, y;
  for (i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  for (i = 1; i <= n; i++) {
    if (v[i].size() == 1) {
      leaves.insert(i);
      tot++;
    }
  }
  dfs(1, 1, 0);
  dfs2(1, -1);
  long long int temp = 0;
  if (leaves.find(1) != leaves.end()) {
    if (v[1].size()) leave[v[1][0]]++;
  }
  for (long long int i = 1; i <= n; i++) {
    if (leave[i]) temp += leave[i] - 1;
  }
  if (odd == tot || even == tot)
    cout << 1 << " ";
  else
    cout << 3 << " ";
  cout << n - 1 - temp << " ";
  return 0;
}

```