#include <bits/stdc++.h>
using namespace std;
const long long int N = 300009;
const long long int mod = 1000000007;
long long int t, n, x, y;
vector<long long int> a[N];
vector<long long int> dp;
void dfs(long long int z, long long int pr) {
  for (auto &it : a[z]) {
    if (it == pr) continue;
    dp[it] = dp[z] + 1;
    dfs(it, z);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin.exceptions(cin.failbit);
  ;
  cin >> n;
  dp.assign(n, 0);
  for (long long int i = 0; i < (n - 1); ++i) {
    cin >> x >> y;
    x--, y--;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  long long int ans = (n - 1);
  long long int l = 0;
  long long int rt = -1;
  for (long long int i = 0; i < (n); ++i) {
    if ((long long int)((a[i]).size()) == 1) {
      l++;
      rt = i;
    }
  }
  for (long long int i = 0; i < (n); ++i) {
    for (auto &it : a[i]) {
      if ((long long int)((a[it]).size()) == 1) {
        l--;
        break;
      }
    }
  }
  dfs(rt, -1);
  for (long long int i = 0; i < (n); ++i) {
    if ((long long int)((a[i]).size()) == 1 and ((dp[i] % 2) == 1)) {
      cout << 3 << " ";
      break;
    }
    if (i == (n - 1)) {
      cout << 1 << " ";
    }
  }
  cout << (ans - l) << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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 int N = 300009;
const long long int mod = 1000000007;
long long int t, n, x, y;
vector<long long int> a[N];
vector<long long int> dp;
void dfs(long long int z, long long int pr) {
  for (auto &it : a[z]) {
    if (it == pr) continue;
    dp[it] = dp[z] + 1;
    dfs(it, z);
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin.exceptions(cin.failbit);
  ;
  cin >> n;
  dp.assign(n, 0);
  for (long long int i = 0; i < (n - 1); ++i) {
    cin >> x >> y;
    x--, y--;
    a[x].push_back(y);
    a[y].push_back(x);
  }
  long long int ans = (n - 1);
  long long int l = 0;
  long long int rt = -1;
  for (long long int i = 0; i < (n); ++i) {
    if ((long long int)((a[i]).size()) == 1) {
      l++;
      rt = i;
    }
  }
  for (long long int i = 0; i < (n); ++i) {
    for (auto &it : a[i]) {
      if ((long long int)((a[it]).size()) == 1) {
        l--;
        break;
      }
    }
  }
  dfs(rt, -1);
  for (long long int i = 0; i < (n); ++i) {
    if ((long long int)((a[i]).size()) == 1 and ((dp[i] % 2) == 1)) {
      cout << 3 << " ";
      break;
    }
    if (i == (n - 1)) {
      cout << 1 << " ";
    }
  }
  cout << (ans - l) << "\n";
  return 0;
}

```