#include <bits/stdc++.h>
#pragma GCC optimize("-Ofast")
using namespace std;
long long power(long long x, long long y, long long m) {
  long long temp;
  if (y == 0) return 1;
  temp = (power(x, y / 2, m)) % m;
  if (y % 2 == 0)
    return ((temp % m) * temp);
  else
    return ((x * temp % m) * temp % m) % m;
}
long long inv(long long x) {
  return (power(x, 1000000007 - 2, 1000000007)) % 1000000007;
}
bool comp(long long a, long long b) { return a > b; }
vector<vector<long long>> edges;
long long visited[1000001], counter = 0;
long long DFS(long long n, long long counter) {
  if (visited[n]) return 1;
  visited[n] = 1;
  long long miner = 1;
  for (long long i = 0; i < edges[n].size(); i++) {
    miner = min(DFS(edges[n][i], counter + 1), miner);
  }
  if (edges[n].size() == 1) return (counter % 2 ? 0 : miner);
  return miner;
}
long long DFS2(long long n) {
  if (visited[n] == 2 && edges[n].size() >= 2)
    return 0;
  else if (visited[n] == 2)
    return 1;
  visited[n] = 2;
  long long flag = 0;
  for (long long i = 0; i < edges[n].size(); i++) {
    if (DFS2(edges[n][i])) flag++;
  }
  if (flag > 1) counter += (flag - 1);
  if (edges[n].size() == 1) return 1;
  return 0;
}
int32_t main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  edges.resize(n);
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    edges[a - 1].push_back(b - 1);
    edges[b - 1].push_back(a - 1);
  }
  long long answer1, answer2;
  for (long long i = 0; i < n; i++) {
    if (edges[i].size() == 1) {
      answer1 = DFS(i, 0);
      DFS2(i);
      break;
    }
  }
  cout << (answer1 ? 1 : 3) << ' ' << (n - 1) - counter;
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
#pragma GCC optimize("-Ofast")
using namespace std;
long long power(long long x, long long y, long long m) {
  long long temp;
  if (y == 0) return 1;
  temp = (power(x, y / 2, m)) % m;
  if (y % 2 == 0)
    return ((temp % m) * temp);
  else
    return ((x * temp % m) * temp % m) % m;
}
long long inv(long long x) {
  return (power(x, 1000000007 - 2, 1000000007)) % 1000000007;
}
bool comp(long long a, long long b) { return a > b; }
vector<vector<long long>> edges;
long long visited[1000001], counter = 0;
long long DFS(long long n, long long counter) {
  if (visited[n]) return 1;
  visited[n] = 1;
  long long miner = 1;
  for (long long i = 0; i < edges[n].size(); i++) {
    miner = min(DFS(edges[n][i], counter + 1), miner);
  }
  if (edges[n].size() == 1) return (counter % 2 ? 0 : miner);
  return miner;
}
long long DFS2(long long n) {
  if (visited[n] == 2 && edges[n].size() >= 2)
    return 0;
  else if (visited[n] == 2)
    return 1;
  visited[n] = 2;
  long long flag = 0;
  for (long long i = 0; i < edges[n].size(); i++) {
    if (DFS2(edges[n][i])) flag++;
  }
  if (flag > 1) counter += (flag - 1);
  if (edges[n].size() == 1) return 1;
  return 0;
}
int32_t main() {
  iostream::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  edges.resize(n);
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    edges[a - 1].push_back(b - 1);
    edges[b - 1].push_back(a - 1);
  }
  long long answer1, answer2;
  for (long long i = 0; i < n; i++) {
    if (edges[i].size() == 1) {
      answer1 = DFS(i, 0);
      DFS2(i);
      break;
    }
  }
  cout << (answer1 ? 1 : 3) << ' ' << (n - 1) - counter;
  return 0;
}

```