#include <bits/stdc++.h>
using namespace std;
vector<int> tree[100005];
bool visited[100005];
int degree[100005];
int N;
bool even = true;
int dfs(int root) {
  if (visited[root]) {
    return 0;
  }
  if (degree[root] == 1) {
    return 1;
  }
  visited[root] = true;
  int sum = 0;
  for (int i : tree[root]) {
    sum |= dfs(i);
  }
  if (sum == 3) return 3;
  return sum == 1 ? 2 : 1;
}
int bfs(int root) {
  queue<int> to_s;
  to_s.push(root);
  int sum = 0;
  while (to_s.size()) {
    int t = to_s.front();
    to_s.pop();
    visited[t] = true;
    int nleave = 0;
    for (int i : tree[t]) {
      if (degree[i] == 1) {
        ++nleave;
        continue;
      }
      if (!visited[i]) {
        ++sum;
        to_s.push(i);
      }
    }
    if (nleave) ++sum;
  }
  return sum;
}
void solve() {
  int root;
  for (root = 1; root < N; ++root) {
    if (degree[root] > 1) break;
  }
  int oddity = dfs(root);
  if (oddity == 3) {
    printf("3 ");
  } else
    printf("1 ");
  memset(visited, 0, sizeof(visited));
  printf("%d", bfs(root));
}
int main() {
  memset(visited, 0, sizeof(visited));
  scanf("%d", &N);
  for (int i = 1; i < N; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    ++degree[u];
    ++degree[v];
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  solve();
}

### Prompt

Create a solution in cpp for 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<int> tree[100005];
bool visited[100005];
int degree[100005];
int N;
bool even = true;
int dfs(int root) {
  if (visited[root]) {
    return 0;
  }
  if (degree[root] == 1) {
    return 1;
  }
  visited[root] = true;
  int sum = 0;
  for (int i : tree[root]) {
    sum |= dfs(i);
  }
  if (sum == 3) return 3;
  return sum == 1 ? 2 : 1;
}
int bfs(int root) {
  queue<int> to_s;
  to_s.push(root);
  int sum = 0;
  while (to_s.size()) {
    int t = to_s.front();
    to_s.pop();
    visited[t] = true;
    int nleave = 0;
    for (int i : tree[t]) {
      if (degree[i] == 1) {
        ++nleave;
        continue;
      }
      if (!visited[i]) {
        ++sum;
        to_s.push(i);
      }
    }
    if (nleave) ++sum;
  }
  return sum;
}
void solve() {
  int root;
  for (root = 1; root < N; ++root) {
    if (degree[root] > 1) break;
  }
  int oddity = dfs(root);
  if (oddity == 3) {
    printf("3 ");
  } else
    printf("1 ");
  memset(visited, 0, sizeof(visited));
  printf("%d", bfs(root));
}
int main() {
  memset(visited, 0, sizeof(visited));
  scanf("%d", &N);
  for (int i = 1; i < N; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    ++degree[u];
    ++degree[v];
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  solve();
}

```