#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 5;
vector<int> g[maxn];
int dep[maxn], du[maxn], qko[maxn];
int n;
int u, v, root;
vector<int> leaf;
void dfs(int x, int par, int d) {
  dep[x] = d;
  if (g[x].size() == 1) {
    leaf.push_back(d);
    qko[par]++;
  }
  for (int son : g[x]) {
    if (son != par) {
      dfs(son, x, d + 1);
    }
  }
}
int r1, r2;
int main() {
  cin >> n;
  r1 = 1, r2 = n - 1;
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    du[u]++, du[v]++;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (du[i] > 1) {
      root = i;
      dfs(i, -1, 0);
      break;
    }
  sort(leaf.begin(), leaf.end());
  for (int i = 0; i <= (int)leaf.size() - 2; i++) {
    if ((leaf[i + 1] - leaf[i]) & 1) {
      r1 = 3;
      break;
    }
  }
  for (int i = 1; i <= n; i++)
    if (qko[i] > 1) r2 -= qko[i] - 1;
  cout << r1 << " " << r2 << "\n";
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 int maxn = 1e6 + 5;
vector<int> g[maxn];
int dep[maxn], du[maxn], qko[maxn];
int n;
int u, v, root;
vector<int> leaf;
void dfs(int x, int par, int d) {
  dep[x] = d;
  if (g[x].size() == 1) {
    leaf.push_back(d);
    qko[par]++;
  }
  for (int son : g[x]) {
    if (son != par) {
      dfs(son, x, d + 1);
    }
  }
}
int r1, r2;
int main() {
  cin >> n;
  r1 = 1, r2 = n - 1;
  for (int i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    du[u]++, du[v]++;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  for (int i = 1; i <= n; i++)
    if (du[i] > 1) {
      root = i;
      dfs(i, -1, 0);
      break;
    }
  sort(leaf.begin(), leaf.end());
  for (int i = 0; i <= (int)leaf.size() - 2; i++) {
    if ((leaf[i + 1] - leaf[i]) & 1) {
      r1 = 3;
      break;
    }
  }
  for (int i = 1; i <= n; i++)
    if (qko[i] > 1) r2 -= qko[i] - 1;
  cout << r1 << " " << r2 << "\n";
}

```