#include <bits/stdc++.h>
using namespace std;
ifstream fin("AAtest.in.txt");
int n, a, b, in, mituvahem, miin = 1;
struct pu {
  int le = 1, klp = 1, klpt = 0, yl, k = 0;
  vector<int> v;
  void mine(pu p[], int &i) {
    while (k != (int)v.size()) {
      if (v[k] != yl) {
        le = 0, klp = 0;
        p[v[k]].yl = i;
        i = v[k];
        k++;
        return;
      } else
        k++;
    }
    k = 0;
    int mv = 0, ps = 0, pt = 0;
    for (int(k) = (0); ((k)) < ((v.size())); ((k))++) {
      if (v[k] == yl) continue;
      if (p[v[k]].le) mv++;
      if (p[v[k]].klp) {
        ps = 1;
        klpt = 1;
      }
      if (p[v[k]].klpt) {
        pt = 1;
        klp = 1;
      }
    }
    if (ps and pt) miin = 3;
    mituvahem += max(mv - 1, 0);
    i = yl;
    return;
  };
} p[100005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n;
  for (int(i) = (0); ((i)) < ((n - 1)); ((i))++) {
    cin >> a >> b;
    p[a].v.emplace_back(b);
    p[b].v.emplace_back(a);
  }
  for (int(i) = (0); ((i)) < ((n + 1)); ((i))++)
    if (i and p[i].v.size() > 1) {
      int in = i;
      p[in].yl = -1;
      while (in != -1) p[in].mine(p, in);
      break;
    }
  cout << miin << " " << n - 1 - mituvahem;
}

### 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;
ifstream fin("AAtest.in.txt");
int n, a, b, in, mituvahem, miin = 1;
struct pu {
  int le = 1, klp = 1, klpt = 0, yl, k = 0;
  vector<int> v;
  void mine(pu p[], int &i) {
    while (k != (int)v.size()) {
      if (v[k] != yl) {
        le = 0, klp = 0;
        p[v[k]].yl = i;
        i = v[k];
        k++;
        return;
      } else
        k++;
    }
    k = 0;
    int mv = 0, ps = 0, pt = 0;
    for (int(k) = (0); ((k)) < ((v.size())); ((k))++) {
      if (v[k] == yl) continue;
      if (p[v[k]].le) mv++;
      if (p[v[k]].klp) {
        ps = 1;
        klpt = 1;
      }
      if (p[v[k]].klpt) {
        pt = 1;
        klp = 1;
      }
    }
    if (ps and pt) miin = 3;
    mituvahem += max(mv - 1, 0);
    i = yl;
    return;
  };
} p[100005];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cerr.tie(0);
  cin >> n;
  for (int(i) = (0); ((i)) < ((n - 1)); ((i))++) {
    cin >> a >> b;
    p[a].v.emplace_back(b);
    p[b].v.emplace_back(a);
  }
  for (int(i) = (0); ((i)) < ((n + 1)); ((i))++)
    if (i and p[i].v.size() > 1) {
      int in = i;
      p[in].yl = -1;
      while (in != -1) p[in].mine(p, in);
      break;
    }
  cout << miin << " " << n - 1 - mituvahem;
}

```