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;
const long long M = 1e9 + 7, N = 205001;
long long mm(long long x, long long y) {
  x %= M, y %= M;
  return (x * y) % M;
}
long long po(long long x, long long y) {
  if (!y) return 1;
  long long a = po(x, y / 2) % M;
  if (y % 2) return mm(a, mm(a, x));
  return mm(a, a);
}
vector<long long> g[N];
bool chk = false;
long long st;
void dfs(long long s, long long p, long long h) {
  long long cnt = 0;
  for (long long i : g[s]) {
    if (i != p) dfs(i, s, h + 1);
    if (g[i].size() == 1) cnt++;
  }
  if (cnt) st++;
  if (g[s].size() == 1 and h % 2) chk = true;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, x, y, l, lc = 0;
  cin >> n;
  for (long long i = 1; 1 < n ? i < n : i > n; 1 < n ? i += 1 : i -= 1) {
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  for (long long i = 1; 1 < n + 1 ? i < n + 1 : i > n + 1;
       1 < n + 1 ? i += 1 : i -= 1)
    if (g[i].size() == 1) {
      l = i;
      lc++;
    }
  dfs(l, 0, 0);
  x = (chk ? 3 : 1);
  y = n - 1 - lc + st;
  cout << x << " " << y;
}
