#include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<vector<long long>> A;
vector<long long> vis;
vector<long long> col;
void DFS(long long v, long long c) {
  col[v] = c;
  vis[v] = +1;
  for (long long i = 0; i < A[v].size(); i++) {
    if (vis[A[v][i]] == 0) {
      DFS(A[v][i], -c);
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long T = 1;
  for (long long iii = 0; iii < T; iii++) {
    long long n;
    cin >> n;
    vector<long long> w;
    vector<long long> deg;
    for (long long i = 0; i <= n; i++) {
      A.push_back(w);
      deg.push_back(0);
      vis.push_back(0);
      col.push_back(0);
    }
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      A[a].push_back(b);
      A[b].push_back(a);
      deg[a]++;
      deg[b]++;
    }
    long long n_out = n;
    for (long long i = 0; i <= n; i++) {
      if (deg[i] == 1) {
        long long b = A[i][0];
        for (long long j = 0; j < deg[b]; j++) {
          if (deg[A[b][j]] == 1 and A[b][j] != i) {
            n_out--;
            deg[A[b][j]]--;
          }
        }
      }
    }
    DFS(1, 1);
    bool flag = false, flag3 = false;
    long long c_tmp;
    for (long long i = 0; i <= n; i++) {
      if (deg[i] == 1) {
        if (flag == false) {
          c_tmp = col[i];
          flag = true;
        } else {
          if (c_tmp != col[i]) {
            flag3 = true;
            break;
          }
        }
      }
    }
    if (flag3 == false) {
      cout << 1 << " " << n_out - 1 << endl;
    } else {
      cout << 3 << " " << n_out - 1 << endl;
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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("sse4")
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
vector<vector<long long>> A;
vector<long long> vis;
vector<long long> col;
void DFS(long long v, long long c) {
  col[v] = c;
  vis[v] = +1;
  for (long long i = 0; i < A[v].size(); i++) {
    if (vis[A[v][i]] == 0) {
      DFS(A[v][i], -c);
    }
  }
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long T = 1;
  for (long long iii = 0; iii < T; iii++) {
    long long n;
    cin >> n;
    vector<long long> w;
    vector<long long> deg;
    for (long long i = 0; i <= n; i++) {
      A.push_back(w);
      deg.push_back(0);
      vis.push_back(0);
      col.push_back(0);
    }
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      A[a].push_back(b);
      A[b].push_back(a);
      deg[a]++;
      deg[b]++;
    }
    long long n_out = n;
    for (long long i = 0; i <= n; i++) {
      if (deg[i] == 1) {
        long long b = A[i][0];
        for (long long j = 0; j < deg[b]; j++) {
          if (deg[A[b][j]] == 1 and A[b][j] != i) {
            n_out--;
            deg[A[b][j]]--;
          }
        }
      }
    }
    DFS(1, 1);
    bool flag = false, flag3 = false;
    long long c_tmp;
    for (long long i = 0; i <= n; i++) {
      if (deg[i] == 1) {
        if (flag == false) {
          c_tmp = col[i];
          flag = true;
        } else {
          if (c_tmp != col[i]) {
            flag3 = true;
            break;
          }
        }
      }
    }
    if (flag3 == false) {
      cout << 1 << " " << n_out - 1 << endl;
    } else {
      cout << 3 << " " << n_out - 1 << endl;
    }
  }
  return 0;
}

```