#include <bits/stdc++.h>
using namespace std;
long long int MAX = LLONG_MAX;
long long int MIN = LLONG_MIN;
const int siz = 200001;
long long int vis[siz];
long long int parent[siz];
vector<long long int> adj[siz];
set<long long int> leaves;
vector<long long int> levels[siz];
long long int P = 1000000007;
long long int flag = 0;
void dfs(long long int src, long long int level) {
  vis[src] = 1;
  levels[level].push_back(src);
  if (leaves.find(src) != leaves.end()) {
    if (level % 2 == 0) flag = 1;
  }
  for (long long int i = 0; i < adj[src].size(); i++) {
    if (!vis[adj[src][i]]) {
      parent[adj[src][i]] = src;
      dfs(adj[src][i], level + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int i, j, k, l, m, n, o, p, q, r, s, t, x, a[500], tt = 0;
  memset(vis, 0, sizeof(vis));
  memset(parent, 0, sizeof(parent));
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    cin >> j >> k;
    adj[j].push_back(k);
    adj[k].push_back(j);
  }
  long long int src = 0;
  for (i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      leaves.insert(i);
      src = i;
    }
  }
  dfs(src, 1);
  long long int mini = 3, maxi = 0;
  if (flag == 0) {
    mini -= 2;
  }
  maxi = n - 1 - leaves.size();
  k = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j < adj[i].size(); j++) {
      if (leaves.find(adj[i][j]) != leaves.end()) {
        k++;
        break;
      }
    }
  }
  maxi += k;
  cout << mini << " " << maxi << "\n";
  ;
  return 0;
}

### Prompt

Your task is to create 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 int MAX = LLONG_MAX;
long long int MIN = LLONG_MIN;
const int siz = 200001;
long long int vis[siz];
long long int parent[siz];
vector<long long int> adj[siz];
set<long long int> leaves;
vector<long long int> levels[siz];
long long int P = 1000000007;
long long int flag = 0;
void dfs(long long int src, long long int level) {
  vis[src] = 1;
  levels[level].push_back(src);
  if (leaves.find(src) != leaves.end()) {
    if (level % 2 == 0) flag = 1;
  }
  for (long long int i = 0; i < adj[src].size(); i++) {
    if (!vis[adj[src][i]]) {
      parent[adj[src][i]] = src;
      dfs(adj[src][i], level + 1);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long int i, j, k, l, m, n, o, p, q, r, s, t, x, a[500], tt = 0;
  memset(vis, 0, sizeof(vis));
  memset(parent, 0, sizeof(parent));
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    cin >> j >> k;
    adj[j].push_back(k);
    adj[k].push_back(j);
  }
  long long int src = 0;
  for (i = 1; i <= n; i++) {
    if (adj[i].size() == 1) {
      leaves.insert(i);
      src = i;
    }
  }
  dfs(src, 1);
  long long int mini = 3, maxi = 0;
  if (flag == 0) {
    mini -= 2;
  }
  maxi = n - 1 - leaves.size();
  k = 0;
  for (i = 1; i <= n; i++) {
    for (j = 0; j < adj[i].size(); j++) {
      if (leaves.find(adj[i][j]) != leaves.end()) {
        k++;
        break;
      }
    }
  }
  maxi += k;
  cout << mini << " " << maxi << "\n";
  ;
  return 0;
}

```