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>
#include <bits/stdc++.h>
using namespace std;
vector<int> al[100005];
int deg[100005];
int dist[100005];
int n;
int fmax() {
  int ans = n - 1;
  for (int u = 1; u <= n; u++) {
    int cnt = 0;
    for (auto v : al[u]) {
      if (deg[v] == 1) cnt++;
    }
    ans -= max(0, cnt - 1);
  }
  return ans;
}
void dfs(int par, int u, int d) {
  dist[u] = d;
  for (auto v : al[u]) {
    if (v != par) dfs(u, v, d + 1);
  }
}
int fmin() {
  dfs(-1, 1, 0);
  int parity = -1;
  for (int u = 1; u <= n; u++) {
    if (deg[u] == 1) {
      if (parity == -1)
        parity = dist[u] % 2;
      else if (dist[u] % 2 != parity)
        return 3;
    }
  }
  return 1;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  scanf("%d", &n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    al[a].push_back(b);
    al[b].push_back(a);
    deg[a]++;
    deg[b]++;
  }
  printf("%d %d", fmin(), fmax());
}
