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 MAXN = 1e5 + 10;
vector<int> g[MAXN];
int dep[MAXN], mx[MAXN];
void dfs(int pos, int par = -1, int d = 0) {
  dep[pos] = d;
  for (auto x : g[pos])
    if (x != par) dfs(x, pos, d + 1);
}
int go(int pos, int par, int d = 1) {
  int ans = d;
  for (auto x : g[pos])
    if (x != par) ans = max(ans, go(x, pos, d + 1));
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1, ThxDem = n; i < ThxDem; ++i) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  queue<int> q;
  vector<int> ds(n, 1e9);
  for (int i = 0, ThxDem = n; i < ThxDem; ++i)
    if (int(g[i].size()) == 1) q.push(i), ds[i] = 0;
  while (int(q.size())) {
    int pos = q.front();
    q.pop();
    for (auto x : g[pos])
      if (ds[x] > ds[pos] + 1) ds[x] = ds[pos] + 1, q.push(pos);
  }
  int rt = 0;
  for (int i = 1, ThxDem = n; i < ThxDem; ++i)
    if (ds[i] > ds[rt]) rt = i;
  dfs(rt);
  {
    int odd = 0, ev = 0;
    for (int i = 0, ThxDem = n; i < ThxDem; ++i)
      if (int(g[i].size()) == 1) odd |= dep[i] % 2 == 1, ev |= dep[i] % 2 == 0;
    if (odd && ev)
      cout << "3 ";
    else
      cout << "1 ";
  }
  {
    int ans = n - 1;
    map<int, int> mp;
    for (int i = 0, ThxDem = n; i < ThxDem; ++i)
      if (int(g[i].size()) == 1) mp[g[i][0]]++;
    for (auto x : mp) ans -= x.second - 1;
    cout << ans << "\n";
  }
}
