#include <bits/stdc++.h>
using namespace std;
long long int m = 1000000007;
vector<long long int> arr[100005];
long long int leafchild[100005] = {};
long long int distt[100005] = {};
long long int visit[100005] = {};
long long int mark[100005] = {};
void dfs(long long int node, long long int dist) {
  visit[node] = 1;
  long long int y = arr[node].size();
  distt[node] = dist;
  for (int i = 0; i < y; i++) {
    if (!visit[arr[node][i]]) {
      long long int yy = arr[arr[node][i]].size();
      if (yy == 1) leafchild[node]++;
      visit[arr[node][i]] = 1;
      dfs(arr[node][i], dist + 1);
    }
    if (!mark[arr[node][i]]) {
      long long int yy = arr[arr[node][i]].size();
      if (yy == 1) leafchild[node]++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int q = 1;
  while (q--) {
    long long int n;
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
      long long int a, b;
      cin >> a >> b;
      arr[a].push_back(b);
      arr[b].push_back(a);
    }
    vector<long long int> leaf;
    for (int i = 1; i <= n; i++) {
      if (arr[i].size() == 1) leaf.push_back(i);
    }
    long long int g = 0;
    long long int min = 0;
    long long int leafnodes = leaf.size();
    long long int sub = 0;
    dfs(1, 0);
    for (int i = 1; i <= n; i++) {
      if (leafchild[i] >= 1) {
        sub++;
      }
    }
    long long int max = n - 1 - leafnodes + sub;
    long long int even = 0;
    long long int odd = 0;
    for (int i = 0; i < leafnodes; i++) {
      if (distt[leaf[i]] % 2 == 0)
        even++;
      else
        odd++;
    }
    if (even > 0 && odd > 0)
      min = 3;
    else
      min = 1;
    cout << min << " " << max;
  }
}

### Prompt

In Cpp, your task is to solve 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 m = 1000000007;
vector<long long int> arr[100005];
long long int leafchild[100005] = {};
long long int distt[100005] = {};
long long int visit[100005] = {};
long long int mark[100005] = {};
void dfs(long long int node, long long int dist) {
  visit[node] = 1;
  long long int y = arr[node].size();
  distt[node] = dist;
  for (int i = 0; i < y; i++) {
    if (!visit[arr[node][i]]) {
      long long int yy = arr[arr[node][i]].size();
      if (yy == 1) leafchild[node]++;
      visit[arr[node][i]] = 1;
      dfs(arr[node][i], dist + 1);
    }
    if (!mark[arr[node][i]]) {
      long long int yy = arr[arr[node][i]].size();
      if (yy == 1) leafchild[node]++;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int q = 1;
  while (q--) {
    long long int n;
    cin >> n;
    for (int i = 0; i < n - 1; i++) {
      long long int a, b;
      cin >> a >> b;
      arr[a].push_back(b);
      arr[b].push_back(a);
    }
    vector<long long int> leaf;
    for (int i = 1; i <= n; i++) {
      if (arr[i].size() == 1) leaf.push_back(i);
    }
    long long int g = 0;
    long long int min = 0;
    long long int leafnodes = leaf.size();
    long long int sub = 0;
    dfs(1, 0);
    for (int i = 1; i <= n; i++) {
      if (leafchild[i] >= 1) {
        sub++;
      }
    }
    long long int max = n - 1 - leafnodes + sub;
    long long int even = 0;
    long long int odd = 0;
    for (int i = 0; i < leafnodes; i++) {
      if (distt[leaf[i]] % 2 == 0)
        even++;
      else
        odd++;
    }
    if (even > 0 && odd > 0)
      min = 3;
    else
      min = 1;
    cout << min << " " << max;
  }
}

```