#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
const int N = 100100;
vector<int> g[N];
int deg[N];
bool good;
int dfs(int v, int p) {
  int rem = -1;
  for (int x : g[v]) {
    if (x == p) continue;
    int tmp = dfs(x, v);
    tmp = (tmp + 1) % 2;
    if (rem == -1) {
      rem = tmp;
      continue;
    }
    if (rem != tmp) {
      good = false;
    }
  }
  if (rem == -1) {
    rem = 0;
  }
  return rem;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int(i) = 0; (i) < (n - 1); (i)++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  int root = -1;
  for (int(i) = 0; (i) < (n); (i)++) {
    if (deg[i] >= 2) {
      root = i;
      break;
    }
  }
  good = true;
  dfs(root, -1);
  if (good)
    cout << "1 ";
  else
    cout << "3 ";
  int res = n - 1;
  for (int(i) = 0; (i) < (n); (i)++) {
    int leaves = 0;
    for (int v : g[i]) {
      if (deg[v] == 1) ++leaves;
    }
    if (leaves >= 2) res -= leaves - 1;
  }
  cout << res << endl;
  return 0;
}

### 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>
#pragma GCC optimize("O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx")
using namespace std;
const int N = 100100;
vector<int> g[N];
int deg[N];
bool good;
int dfs(int v, int p) {
  int rem = -1;
  for (int x : g[v]) {
    if (x == p) continue;
    int tmp = dfs(x, v);
    tmp = (tmp + 1) % 2;
    if (rem == -1) {
      rem = tmp;
      continue;
    }
    if (rem != tmp) {
      good = false;
    }
  }
  if (rem == -1) {
    rem = 0;
  }
  return rem;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  for (int(i) = 0; (i) < (n - 1); (i)++) {
    int a, b;
    cin >> a >> b;
    --a, --b;
    g[a].push_back(b);
    g[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  int root = -1;
  for (int(i) = 0; (i) < (n); (i)++) {
    if (deg[i] >= 2) {
      root = i;
      break;
    }
  }
  good = true;
  dfs(root, -1);
  if (good)
    cout << "1 ";
  else
    cout << "3 ";
  int res = n - 1;
  for (int(i) = 0; (i) < (n); (i)++) {
    int leaves = 0;
    for (int v : g[i]) {
      if (deg[v] == 1) ++leaves;
    }
    if (leaves >= 2) res -= leaves - 1;
  }
  cout << res << endl;
  return 0;
}

```