#include <bits/stdc++.h>
using namespace std;
vector<vector<int> > edge;
vector<int> leave;
vector<int> depth;
vector<int> xet;
void dfs(int root, int d) {
  depth[root] = d;
  xet[root] = 1;
  for (int i = 0; i < edge[root].size(); i++) {
    if (xet[edge[root][i]] == 1)
      continue;
    else
      dfs(edge[root][i], d + 1);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  edge.resize(n + 1);
  depth.resize(n + 1);
  xet.resize(n + 1);
  for (int i = 0; i < n + 1; i++) {
    xet[i] = 0;
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  int root;
  for (int i = 1; i <= n; i++) {
    if (edge[i].size() == 1) {
      root = i;
      leave.push_back(edge[i][0]);
    }
  }
  sort(leave.begin(), leave.end());
  int count = 0;
  for (int i = 1; i < leave.size(); i++) {
    if (leave[i] == leave[i - 1]) count++;
  }
  dfs(root, 0);
  int check = 0;
  for (int i = 1; i <= n; i++) {
    if (depth[i] % 2 == 1 && edge[i].size() == 1) {
      check = 1;
      break;
    }
  }
  int min, max;
  if (check == 0)
    min = 1;
  else
    min = 3;
  max = n - 1 - count;
  printf("%d %d \n", min, max);
}

### Prompt

Please create a solution in CPP 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;
vector<vector<int> > edge;
vector<int> leave;
vector<int> depth;
vector<int> xet;
void dfs(int root, int d) {
  depth[root] = d;
  xet[root] = 1;
  for (int i = 0; i < edge[root].size(); i++) {
    if (xet[edge[root][i]] == 1)
      continue;
    else
      dfs(edge[root][i], d + 1);
  }
}
int main() {
  int n;
  scanf("%d", &n);
  edge.resize(n + 1);
  depth.resize(n + 1);
  xet.resize(n + 1);
  for (int i = 0; i < n + 1; i++) {
    xet[i] = 0;
  }
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    edge[a].push_back(b);
    edge[b].push_back(a);
  }
  int root;
  for (int i = 1; i <= n; i++) {
    if (edge[i].size() == 1) {
      root = i;
      leave.push_back(edge[i][0]);
    }
  }
  sort(leave.begin(), leave.end());
  int count = 0;
  for (int i = 1; i < leave.size(); i++) {
    if (leave[i] == leave[i - 1]) count++;
  }
  dfs(root, 0);
  int check = 0;
  for (int i = 1; i <= n; i++) {
    if (depth[i] % 2 == 1 && edge[i].size() == 1) {
      check = 1;
      break;
    }
  }
  int min, max;
  if (check == 0)
    min = 1;
  else
    min = 3;
  max = n - 1 - count;
  printf("%d %d \n", min, max);
}

```