#include <bits/stdc++.h>
using namespace std;
const long long N = 100005;
const long long M = 1000000007;
long long a[N];
void dfs(long long st, vector<vector<long long> > &v, vector<bool> &vis) {
  if (vis[st]) return;
  vis[st] = true;
  for (long long i = 0; i < v[st].size(); i++) {
    if (vis[v[st][i]]) continue;
    a[v[st][i]] = 1 + a[st];
    dfs(v[st][i], v, vis);
  }
  return;
}
long long pow1(long long a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % M;
    }
    a = (a * a) % M;
    b >>= 1;
  }
  return res % M;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<vector<long long> > v(n + 1);
  vector<bool> vis(n + 1, false);
  long long x, y;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  if (n == 2) {
    cout << 1 << " " << 1;
    return 0;
  }
  bool od = false;
  bool evn = false;
  dfs(1, v, vis);
  for (long long i = 1; i < n + 1; i++) {
    if (v[i].size() == 1 && a[i] % 2 == 1) od = true;
    if (v[i].size() == 1 && a[i] % 2 == 0) evn = true;
  }
  if (od && evn)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  long long ans = n - 1;
  for (long long i = 1; i < n + 1; i++) {
    long long same = 0;
    for (long long j = 0; j < v[i].size(); j++) {
      if (v[v[i][j]].size() == 1) same++;
    }
    if (same > 1) ans -= same - 1;
  }
  cout << ans;
  return 0;
}

### Prompt

Please formulate a CPP solution to 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 long long N = 100005;
const long long M = 1000000007;
long long a[N];
void dfs(long long st, vector<vector<long long> > &v, vector<bool> &vis) {
  if (vis[st]) return;
  vis[st] = true;
  for (long long i = 0; i < v[st].size(); i++) {
    if (vis[v[st][i]]) continue;
    a[v[st][i]] = 1 + a[st];
    dfs(v[st][i], v, vis);
  }
  return;
}
long long pow1(long long a, long long b) {
  int res = 1;
  while (b > 0) {
    if (b & 1) {
      res = (res * a) % M;
    }
    a = (a * a) % M;
    b >>= 1;
  }
  return res % M;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<vector<long long> > v(n + 1);
  vector<bool> vis(n + 1, false);
  long long x, y;
  for (long long i = 0; i < n - 1; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  if (n == 2) {
    cout << 1 << " " << 1;
    return 0;
  }
  bool od = false;
  bool evn = false;
  dfs(1, v, vis);
  for (long long i = 1; i < n + 1; i++) {
    if (v[i].size() == 1 && a[i] % 2 == 1) od = true;
    if (v[i].size() == 1 && a[i] % 2 == 0) evn = true;
  }
  if (od && evn)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  long long ans = n - 1;
  for (long long i = 1; i < n + 1; i++) {
    long long same = 0;
    for (long long j = 0; j < v[i].size(); j++) {
      if (v[v[i][j]].size() == 1) same++;
    }
    if (same > 1) ans -= same - 1;
  }
  cout << ans;
  return 0;
}

```