#include <bits/stdc++.h>
using namespace std;
;
const int max_n = 1e5 + 100;
int n;
bool vis[max_n];
vector<int> T[max_n];
struct node {
  int from = 0;
  vector<int> next;
} G[max_n];
int cmp;
bool judge(int root) {
  if (G[root].next.size() == 0) {
    cmp = 1;
    return true;
  }
  int now_cmp = -1;
  for (int child = 0; child < G[root].next.size(); child++) {
    if (!judge(G[root].next[child])) return false;
    if (now_cmp == -1) {
      now_cmp = cmp;
      if (now_cmp == 1 && G[root].from == 0 && G[root].next.size() == 1)
        return false;
    } else if (cmp != now_cmp)
      return false;
  }
  cmp = now_cmp ^ 1;
  return true;
}
void bfs(int start) {
  deque<int> que;
  que.push_back(start);
  while (!que.empty()) {
    int cur = que.front();
    que.pop_front();
    vis[cur] = true;
    for (int i = 0; i < T[cur].size(); i++) {
      if (vis[T[cur][i]]) continue;
      G[cur].next.push_back(T[cur][i]);
      G[T[cur][i]].from = cur;
      que.push_back(T[cur][i]);
    }
  }
  fill(vis, vis + n + 1, false);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    T[a].push_back(b);
    T[b].push_back(a);
  }
  bfs(1);
  if (judge(1))
    cout << 1 << " ";
  else
    cout << 3 << " ";
  int leaves = 0;
  int above = 0;
  for (int i = 2; i <= n; i++) {
    if (G[i].next.size() == 0) {
      leaves++;
      if (vis[G[i].from]) continue;
      vis[G[i].from] = true;
      above++;
    }
  }
  if (G[1].next.size() == 1) {
    leaves++;
    if (!vis[G[1].next[0]]) above++;
  }
  cout << n - leaves + above - 1 << endl;
}

### 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;
;
const int max_n = 1e5 + 100;
int n;
bool vis[max_n];
vector<int> T[max_n];
struct node {
  int from = 0;
  vector<int> next;
} G[max_n];
int cmp;
bool judge(int root) {
  if (G[root].next.size() == 0) {
    cmp = 1;
    return true;
  }
  int now_cmp = -1;
  for (int child = 0; child < G[root].next.size(); child++) {
    if (!judge(G[root].next[child])) return false;
    if (now_cmp == -1) {
      now_cmp = cmp;
      if (now_cmp == 1 && G[root].from == 0 && G[root].next.size() == 1)
        return false;
    } else if (cmp != now_cmp)
      return false;
  }
  cmp = now_cmp ^ 1;
  return true;
}
void bfs(int start) {
  deque<int> que;
  que.push_back(start);
  while (!que.empty()) {
    int cur = que.front();
    que.pop_front();
    vis[cur] = true;
    for (int i = 0; i < T[cur].size(); i++) {
      if (vis[T[cur][i]]) continue;
      G[cur].next.push_back(T[cur][i]);
      G[T[cur][i]].from = cur;
      que.push_back(T[cur][i]);
    }
  }
  fill(vis, vis + n + 1, false);
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    T[a].push_back(b);
    T[b].push_back(a);
  }
  bfs(1);
  if (judge(1))
    cout << 1 << " ";
  else
    cout << 3 << " ";
  int leaves = 0;
  int above = 0;
  for (int i = 2; i <= n; i++) {
    if (G[i].next.size() == 0) {
      leaves++;
      if (vis[G[i].from]) continue;
      vis[G[i].from] = true;
      above++;
    }
  }
  if (G[1].next.size() == 1) {
    leaves++;
    if (!vis[G[1].next[0]]) above++;
  }
  cout << n - leaves + above - 1 << endl;
}

```