#include <bits/stdc++.h>
using namespace std;
long long INF = 1e9 + 1;
vector<long long> dist;
vector<long long> used;
vector<vector<long long>> g;
long long x = -1;
void dfs(long long v, long long p) {
  used[v] = true;
  if (v != x) dist[v] = dist[p] + 1;
  for (auto &u : g[v]) {
    if (!used[u]) dfs(u, v);
  }
}
signed main() {
  long long n;
  cin >> n;
  dist.resize(n);
  used.resize(n);
  g.resize(n);
  for (long long i = 1; i < n; ++i) {
    long long a, b;
    cin >> a >> b;
    g[a - 1].push_back(b - 1);
    g[b - 1].push_back(a - 1);
  }
  long long leafs = 0;
  vector<bool> le(n);
  for (long long i = 0; i < n; ++i) {
    if (g[i].size() > 1) {
      x = i;
    } else {
      le[i] = true;
      ++leafs;
    }
  }
  dist[x] = 0;
  dfs(x, x);
  long long ch = 0, nch = 0;
  for (long long i = 0; i < n; ++i) {
    if (le[i]) {
      if ((dist[i] % 2) == 0)
        ch = 1;
      else
        nch = 1;
    }
  }
  if (ch != nch) {
    cout << 1 << ' ';
  } else {
    cout << 3 << ' ';
  }
  long long fe = 0;
  vector<bool> fe_l(n);
  for (long long i = 0; i < n; ++i) {
    if (le[i]) {
      fe_l[g[i][0]] = true;
    }
  }
  for (long long i = 0; i < n; ++i) {
    if (fe_l[i]) {
      ++fe;
    }
  }
  cout << n - 1 - leafs + fe;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
long long INF = 1e9 + 1;
vector<long long> dist;
vector<long long> used;
vector<vector<long long>> g;
long long x = -1;
void dfs(long long v, long long p) {
  used[v] = true;
  if (v != x) dist[v] = dist[p] + 1;
  for (auto &u : g[v]) {
    if (!used[u]) dfs(u, v);
  }
}
signed main() {
  long long n;
  cin >> n;
  dist.resize(n);
  used.resize(n);
  g.resize(n);
  for (long long i = 1; i < n; ++i) {
    long long a, b;
    cin >> a >> b;
    g[a - 1].push_back(b - 1);
    g[b - 1].push_back(a - 1);
  }
  long long leafs = 0;
  vector<bool> le(n);
  for (long long i = 0; i < n; ++i) {
    if (g[i].size() > 1) {
      x = i;
    } else {
      le[i] = true;
      ++leafs;
    }
  }
  dist[x] = 0;
  dfs(x, x);
  long long ch = 0, nch = 0;
  for (long long i = 0; i < n; ++i) {
    if (le[i]) {
      if ((dist[i] % 2) == 0)
        ch = 1;
      else
        nch = 1;
    }
  }
  if (ch != nch) {
    cout << 1 << ' ';
  } else {
    cout << 3 << ' ';
  }
  long long fe = 0;
  vector<bool> fe_l(n);
  for (long long i = 0; i < n; ++i) {
    if (le[i]) {
      fe_l[g[i][0]] = true;
    }
  }
  for (long long i = 0; i < n; ++i) {
    if (fe_l[i]) {
      ++fe;
    }
  }
  cout << n - 1 - leafs + fe;
}

```