#include <bits/stdc++.h>
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const int N = 100005;
int n, a, b;
vector<int> V[N];
bool isLeaf[2];
int cntBad;
bool used[N];
void DFS(int w, int d, int par = -1) {
  if (V[w].size() == 1) {
    isLeaf[d] = true;
    if (used[par]) {
      cntBad++;
    }
    used[par] = true;
  }
  for (int u : V[w]) {
    if (par == u) {
      continue;
    }
    DFS(u, d ^ 1, w);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d %d", &a, &b);
    V[a].push_back(b);
    V[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (V[i].size() >= 2) {
      DFS(i, 0);
      break;
    }
  }
  int mini = (!isLeaf[0] || !isLeaf[1]) ? 1 : 3;
  int maxi = n - cntBad - 1;
  printf("%d %d\n", mini, maxi);
  return 0;
}

### 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;
template <class TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
  while (*sdbg != ',') cerr << *sdbg++;
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, a...);
}
template <class T>
ostream &operator<<(ostream &os, vector<T> V) {
  os << "[";
  for (auto vv : V) os << vv << ",";
  return os << "]";
}
template <class L, class R>
ostream &operator<<(ostream &os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
const int N = 100005;
int n, a, b;
vector<int> V[N];
bool isLeaf[2];
int cntBad;
bool used[N];
void DFS(int w, int d, int par = -1) {
  if (V[w].size() == 1) {
    isLeaf[d] = true;
    if (used[par]) {
      cntBad++;
    }
    used[par] = true;
  }
  for (int u : V[w]) {
    if (par == u) {
      continue;
    }
    DFS(u, d ^ 1, w);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    scanf("%d %d", &a, &b);
    V[a].push_back(b);
    V[b].push_back(a);
  }
  for (int i = 1; i <= n; i++) {
    if (V[i].size() >= 2) {
      DFS(i, 0);
      break;
    }
  }
  int mini = (!isLeaf[0] || !isLeaf[1]) ? 1 : 3;
  int maxi = n - cntBad - 1;
  printf("%d %d\n", mini, maxi);
  return 0;
}

```