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;
template <typename T = int>
T input() {
  T res;
  cin >> res;
  return res;
}
constexpr int N = 1e5 + 7;
vector<pair<int, int> > g[N], leafs[N];
int pr[N], comp;
bool odd = false, even = false;
int rt(int x) {
  if (pr[x] == x) {
    return x;
  }
  return pr[x] = rt(pr[x]);
}
void merge(int x, int y) {
  int rx = rt(x), ry = rt(y);
  if (rand() % 2) {
    swap(rx, ry);
  }
  if (rx != ry) {
    pr[rx] = ry;
    --comp;
  }
}
void dfs(int v, int p, int d) {
  if ((int)g[v].size() == 1) {
    if (d % 2) {
      odd = true;
    } else {
      even = true;
    }
  }
  for (auto [u, ind] : g[v]) {
    if (u != p) {
      dfs(u, v, d + 1);
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n = input();
  iota(pr, pr + n, 0);
  comp = n - 1;
  for (int i = 0; i < n - 1; ++i) {
    int a = input() - 1;
    int b = input() - 1;
    g[a].emplace_back(b, i);
    g[b].emplace_back(a, i);
  }
  for (int i = 0; i < n; ++i) {
    if ((int)g[i].size() == 1) {
      for (auto [u, ind] : g[i]) {
        leafs[u].emplace_back(i, ind);
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    if ((int)leafs[i].size() > 1) {
      int fr = leafs[i].front().second;
      for (auto [u, ind] : leafs[i]) {
        merge(ind, fr);
      }
    }
  }
  dfs(0, -1, 0);
  cout << (odd && even ? 3 : 1) << ' ' << comp << endl;
}
