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 bool ready = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(12);
  return true;
}();
const double PI = acos(-1);
using ll = long long;
using pii = pair<ll, ll>;
using pdd = pair<double, double>;
using vd = vector<double>;
using vb = vector<bool>;
using vi = vector<ll>;
using vvi = vector<vi>;
using vs = vector<string>;
void solve() {
  ll n;
  cin >> n;
  ;
  vvi adj(n);
  for (ll i = 0; i < n - 1; i++) {
    ll a;
    cin >> a;
    ;
    a--;
    ll b;
    cin >> b;
    ;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  vi d(n);
  function<void(ll, ll)> dfs = [&](ll v, ll p) {
    if (p >= 0)
      d[v] = d[p] + 1;
    else
      d[v] = 0;
    for (auto chl : adj[v]) {
      if (chl != p) {
        dfs(chl, v);
      }
    }
  };
  for (ll i = 0; i < n; i++) {
    if (adj[i].size() == 1) {
      dfs(i, -1);
      break;
    }
  }
  ll hasOdd = false;
  for (ll i = 0; i < n; i++)
    if (adj[i].size() == 1 && d[i] % 2 == 1) hasOdd = true;
  ll ansmi = 0;
  if (!hasOdd)
    ansmi = 1;
  else
    ansmi = 3;
  ll ansma = 0;
  for (ll i = 0; i < n; i++) {
    if (adj[i].size() > 1) {
      ll cnt = 0;
      for (ll c : adj[i])
        if (adj[c].size() == 1) cnt++;
      ansma += max(0LL, cnt - 1);
    }
  }
  cout << ansmi << " " << n - 1 - ansma << endl;
}
signed main() { solve(); }
