#include <bits/stdc++.h>
using namespace std;
int n, from, to, minr, maxr, u[100014], d[100014], root, ceven, codd;
vector<int> g[100014];
set<int> leaves;
void go(int cx, int depth) {
  if (u[cx]) return;
  u[cx] = 1;
  d[cx] = depth;
  for (int i = (0); i < (g[cx].size()); i++) go(g[cx][i], depth + 1);
}
void gomax(int cx) {
  if (u[cx]) return;
  u[cx] = 1;
  int lcnt = 0;
  for (int i = (0); i < (g[cx].size()); i++) {
    if (g[g[cx][i]].size() == 1) ++lcnt;
    gomax(g[cx][i]);
  }
  if (lcnt >= 2) maxr -= (lcnt - 1);
}
int main() {
  do {
    scanf("%d", &(n));
  } while (0);
  for (int i = (0); i < (n - 1); i++) {
    do {
      scanf("%d%d", &(from), &(to));
    } while (0);
    --from;
    --to;
    g[from].push_back(to);
    g[to].push_back(from);
  }
  for (int i = (0); i < (n); i++)
    if (g[i].size() == 1) leaves.insert(i);
  root = *leaves.begin();
  memset((u), (0), sizeof((u)));
  go(root, 0);
  ceven = codd = 0;
  for (int i = (0); i < (n); i++)
    if (leaves.count(i)) {
      if (d[i] % 2)
        ++ceven;
      else
        ++codd;
    }
  if (ceven && codd)
    minr = 3;
  else
    minr = 1;
  maxr = n - 1;
  memset((u), (0), sizeof((u)));
  gomax(0);
  printf("%d %d\n", minr, maxr);
  return 0;
}

### Prompt

Generate 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;
int n, from, to, minr, maxr, u[100014], d[100014], root, ceven, codd;
vector<int> g[100014];
set<int> leaves;
void go(int cx, int depth) {
  if (u[cx]) return;
  u[cx] = 1;
  d[cx] = depth;
  for (int i = (0); i < (g[cx].size()); i++) go(g[cx][i], depth + 1);
}
void gomax(int cx) {
  if (u[cx]) return;
  u[cx] = 1;
  int lcnt = 0;
  for (int i = (0); i < (g[cx].size()); i++) {
    if (g[g[cx][i]].size() == 1) ++lcnt;
    gomax(g[cx][i]);
  }
  if (lcnt >= 2) maxr -= (lcnt - 1);
}
int main() {
  do {
    scanf("%d", &(n));
  } while (0);
  for (int i = (0); i < (n - 1); i++) {
    do {
      scanf("%d%d", &(from), &(to));
    } while (0);
    --from;
    --to;
    g[from].push_back(to);
    g[to].push_back(from);
  }
  for (int i = (0); i < (n); i++)
    if (g[i].size() == 1) leaves.insert(i);
  root = *leaves.begin();
  memset((u), (0), sizeof((u)));
  go(root, 0);
  ceven = codd = 0;
  for (int i = (0); i < (n); i++)
    if (leaves.count(i)) {
      if (d[i] % 2)
        ++ceven;
      else
        ++codd;
    }
  if (ceven && codd)
    minr = 3;
  else
    minr = 1;
  maxr = n - 1;
  memset((u), (0), sizeof((u)));
  gomax(0);
  printf("%d %d\n", minr, maxr);
  return 0;
}

```