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;
char buf[1 << 21], *p1 = buf, *p2 = buf, obuf[1 << 23], *O = obuf;
inline int read() {
  int x = 0, sign = 0;
  char s = getchar();
  while (!isdigit(s)) sign |= s == '-', s = getchar();
  while (isdigit(s)) x = (x << 1) + (x << 3) + (s - '0'), s = getchar();
  return sign ? -x : x;
}
void print(int x) {
  if (x > 9) print(x / 10);
  (*O++ = x % 10 + '0');
}
const int N = 3e5 + 5;
int n, l[N], fa[N], ma;
vector<int> e[N], d1[N], d2[N], dep[2];
void dfs(int id, int f, int d) {
  fa[id] = f;
  if (l[id])
    d1[fa[id]].push_back(id), d2[fa[fa[id]]].push_back(id),
        dep[d & 1].push_back(id);
  for (auto it : e[id])
    if (it != f) dfs(it, id, d + 1);
  if (d1[id].size()) ma -= d1[id].size() - 1;
}
void solve() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    e[u].push_back(v), e[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (e[i].size() == 1) l[i] = 1;
  for (int i = 1; i <= n; i++)
    if (e[i].size() > 1) {
      ma = n - 1, dfs(i, 0, 0);
      cout << (dep[0].size() && dep[1].size() ? 3 : 1) << " " << ma << endl;
      return;
    }
}
int main() {
  int testcase = 1;
  while (testcase--) solve();
  return 0;
}
