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;
int cnt, n, ans, to[200005], nxt[200005], h[100005], deg[100005];
bool f[100005][2];
void add(int s, int t) {
  to[++cnt] = t;
  nxt[cnt] = h[s];
  h[s] = cnt;
}
void dfs1(int st, int fa = 0) {
  if (!nxt[h[st]]) f[st][0] = true;
  for (int i = h[st]; i; i = nxt[i])
    if (to[i] != fa) {
      dfs1(to[i], st);
      if (f[to[i]][0] && f[st][0]) ans = 3;
      if (f[to[i]][1] && f[st][1]) ans = 3;
      f[st][0] |= f[to[i]][1];
      f[st][1] |= f[to[i]][0];
    }
}
void dfs2(int st, int fa = 0) {
  if (!nxt[h[st]]) {
    if (st == 1)
      deg[to[h[st]]]++;
    else {
      deg[fa]++;
      return;
    }
  }
  for (int i = h[st]; i; i = nxt[i])
    if (to[i] != fa) dfs2(to[i], st);
}
int main() {
  int i, s, t;
  cin >> n;
  for (i = 1; i < n; i++) {
    cin >> s >> t;
    add(s, t);
    add(t, s);
  }
  ans = 1;
  dfs1(1);
  cout << ans << " ";
  ans = n - 1;
  dfs2(1);
  for (i = 1; i <= n; i++)
    if (deg[i]) ans -= deg[i] - 1;
  cout << ans << endl;
  return 0;
}
