#include <bits/stdc++.h>
using namespace std;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int ddx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, ddy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
void mad(long long &a, long long b) {
  a = (a + b) % 1000000007;
  if (a < 0) a += 1000000007;
}
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int n;
int ans;
int d[100005];
vector<int> ed[100005];
void dfs(int i, int h, int p) {
  d[i] = h;
  int cnt = 0;
  for (int j : ed[i]) {
    if (j == p) continue;
    if (ed[j].size() == 1) cnt++;
    dfs(j, h + 1, i);
  }
  if (cnt) ans = ans - cnt + 1;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    ed[u].push_back(v);
    ed[v].push_back(u);
  }
  int root = 1;
  for (int i = 1; i <= n; i++)
    if (ed[i].size() > 1) {
      root = i;
      break;
    }
  dfs(root, 0, 0);
  int e = 0, o = 0;
  for (int i = 1; i <= n; i++) {
    if (ed[i].size() == 1) {
      if (d[i] & 1)
        o = 1;
      else
        e = 1;
    }
  }
  if (e & o)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  ans += n - 1;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Your task is to create 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;
int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
int ddx[8] = {1, 1, 0, -1, -1, -1, 0, 1}, ddy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
void mad(long long &a, long long b) {
  a = (a + b) % 1000000007;
  if (a < 0) a += 1000000007;
}
long long gcd(long long a, long long b) {
  if (!a) return b;
  return gcd(b % a, a);
}
int n;
int ans;
int d[100005];
vector<int> ed[100005];
void dfs(int i, int h, int p) {
  d[i] = h;
  int cnt = 0;
  for (int j : ed[i]) {
    if (j == p) continue;
    if (ed[j].size() == 1) cnt++;
    dfs(j, h + 1, i);
  }
  if (cnt) ans = ans - cnt + 1;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    ed[u].push_back(v);
    ed[v].push_back(u);
  }
  int root = 1;
  for (int i = 1; i <= n; i++)
    if (ed[i].size() > 1) {
      root = i;
      break;
    }
  dfs(root, 0, 0);
  int e = 0, o = 0;
  for (int i = 1; i <= n; i++) {
    if (ed[i].size() == 1) {
      if (d[i] & 1)
        o = 1;
      else
        e = 1;
    }
  }
  if (e & o)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  ans += n - 1;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}

```