#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 != 0) {
      ans = (ans * a) % 1000000007;
    }
    a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
    b >>= 1;
  }
  return ans;
}
void dfs1(vector<int> a[], int i, bool vis[], int height, bool& flag, bool& f) {
  vis[i] = true;
  if (a[i].size() == 1 && height % 2 != 0) flag = true;
  if (a[i].size() == 1 && height % 2 == 0) f = true;
  for (auto j : a[i]) {
    if (vis[j] == true) continue;
    dfs1(a, j, vis, height + 1, flag, f);
  }
}
int dfs(int i, vector<int> a[], bool vis[]) {
  vis[i] = true;
  vector<int> v;
  int val = 0, z = -1, cnt = 0;
  for (auto j : a[i]) {
    if (vis[j] == true) continue;
    z = dfs(j, a, vis);
    if (z == 0) cnt++;
    val += z + 1;
  }
  if (z == -1) return 0;
  if (cnt != 0) val -= (cnt - 1);
  ;
  return val;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a[n + 1];
    int root = -1;
    for (int i = 0; i < n - 1; i++) {
      int p, q;
      cin >> p >> q;
      a[p].push_back(q);
      a[q].push_back(p);
      if (a[p].size() > 1) root = p;
      if (a[q].size() > 1) root = q;
    }
    bool flag = false, f = false;
    bool vis[n + 1];
    memset(vis, false, sizeof(vis));
    dfs1(a, root, vis, 0, flag, f);
    if (flag == true && f == true)
      cout << "3 ";
    else
      cout << "1 ";
    memset(vis, false, sizeof(vis));
    cout << dfs(root, a, vis);
  }
  return 0;
}

### Prompt

Please formulate 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;
long long power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 != 0) {
      ans = (ans * a) % 1000000007;
    }
    a = ((a % 1000000007) * (a % 1000000007)) % 1000000007;
    b >>= 1;
  }
  return ans;
}
void dfs1(vector<int> a[], int i, bool vis[], int height, bool& flag, bool& f) {
  vis[i] = true;
  if (a[i].size() == 1 && height % 2 != 0) flag = true;
  if (a[i].size() == 1 && height % 2 == 0) f = true;
  for (auto j : a[i]) {
    if (vis[j] == true) continue;
    dfs1(a, j, vis, height + 1, flag, f);
  }
}
int dfs(int i, vector<int> a[], bool vis[]) {
  vis[i] = true;
  vector<int> v;
  int val = 0, z = -1, cnt = 0;
  for (auto j : a[i]) {
    if (vis[j] == true) continue;
    z = dfs(j, a, vis);
    if (z == 0) cnt++;
    val += z + 1;
  }
  if (z == -1) return 0;
  if (cnt != 0) val -= (cnt - 1);
  ;
  return val;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a[n + 1];
    int root = -1;
    for (int i = 0; i < n - 1; i++) {
      int p, q;
      cin >> p >> q;
      a[p].push_back(q);
      a[q].push_back(p);
      if (a[p].size() > 1) root = p;
      if (a[q].size() > 1) root = q;
    }
    bool flag = false, f = false;
    bool vis[n + 1];
    memset(vis, false, sizeof(vis));
    dfs1(a, root, vis, 0, flag, f);
    if (flag == true && f == true)
      cout << "3 ";
    else
      cout << "1 ";
    memset(vis, false, sizeof(vis));
    cout << dfs(root, a, vis);
  }
  return 0;
}

```