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 int mod = 998244353, N = 110000;
int n, dis[N], flag = 1, vis[N];
vector<int> to[N];
inline void addEdg(int x, int y) {
  to[x].push_back(y), to[y].push_back(x);
  return;
}
inline int addMod(int a, int b) { return (a += b) >= mod ? a - mod : a; }
inline long long quickpow(long long base, long long pw) {
  long long ret = 1;
  while (pw) {
    if (pw & 1) ret = ret * base % mod;
    base = base * base % mod, pw >>= 1;
  }
  return ret;
}
template <class T>
inline void read(T &x) {
  x = 0;
  char ch = getchar(), w = 0;
  while (!isdigit(ch)) w = (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = (x << 1) + (x << 3) + (ch ^ 48), ch = getchar();
  x = w ? -x : x;
  return;
}
void dfs(int now, int fa) {
  for (auto &v : to[now]) {
    if (v == fa) continue;
    dis[v] = dis[now] + 1, dfs(v, now);
  }
  return;
}
int main() {
  int x, y;
  read(n);
  for (register int i = 1; i < n; ++i) read(x), read(y), addEdg(x, y);
  dfs(1, 0);
  int ans = n - 1, f0 = 0, f1 = 0;
  for (register int i = 1; i <= n; ++i) {
    if (to[i].size() != 1) continue;
    if (dis[i] & 1)
      f1 = 1;
    else
      f0 = 1;
    if (vis[to[i][0]]) --ans;
    vis[to[i][0]] = 1;
  }
  if (f0 && f1) flag = 3;
  cout << flag << ' ' << ans << endl;
  return 0;
}
