#include <bits/stdc++.h>
using namespace std;
const long long MXN = 2e5 + 10;
long long n, maxi, t, _Min, _Max;
long long d[MXN], o[MXN], e[MXN];
set<long long> st;
vector<long long> adj[MXN];
void dfs(long long u, long long p) {
  for (auto v : adj[u]) {
    if (v == p) continue;
    dfs(v, u);
    if (d[v] == 1) o[u]++;
    o[u] += e[v];
    e[u] += o[v];
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v), adj[v].push_back(u);
    d[u]++, d[v]++;
    maxi = max({maxi, d[u], d[v]});
  }
  if (n == 3) {
    cout << 1 << ' ' << 1;
    return 0;
  }
  if (maxi == 2) {
    if ((n - 1) % 2)
      cout << 3 << ' ' << n - 1;
    else
      cout << 1 << ' ' << n - 1;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (d[i] > 1) continue;
    st.insert(adj[i][0]);
    t++;
  }
  t = t - st.size();
  _Max = n - 1 - t, _Min = 1;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    if (o[i] && e[i]) _Min = 3;
  }
  cout << _Min << ' ' << _Max << '\n';
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long long MXN = 2e5 + 10;
long long n, maxi, t, _Min, _Max;
long long d[MXN], o[MXN], e[MXN];
set<long long> st;
vector<long long> adj[MXN];
void dfs(long long u, long long p) {
  for (auto v : adj[u]) {
    if (v == p) continue;
    dfs(v, u);
    if (d[v] == 1) o[u]++;
    o[u] += e[v];
    e[u] += o[v];
  }
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v), adj[v].push_back(u);
    d[u]++, d[v]++;
    maxi = max({maxi, d[u], d[v]});
  }
  if (n == 3) {
    cout << 1 << ' ' << 1;
    return 0;
  }
  if (maxi == 2) {
    if ((n - 1) % 2)
      cout << 3 << ' ' << n - 1;
    else
      cout << 1 << ' ' << n - 1;
    return 0;
  }
  for (int i = 1; i <= n; i++) {
    if (d[i] > 1) continue;
    st.insert(adj[i][0]);
    t++;
  }
  t = t - st.size();
  _Max = n - 1 - t, _Min = 1;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    if (o[i] && e[i]) _Min = 3;
  }
  cout << _Min << ' ' << _Max << '\n';
  return 0;
}

```