#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long ppow(long long x, long long y, long long mod) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return res;
}
const int N = 1e5 + 5;
vector<vector<int>> g;
int ans1 = 1;
void dfs(int v, int pr, int d) {
  if (g[v].size() == 1 && d % 2) {
    ans1 = 3;
    return;
  }
  for (int to : g[v]) {
    if (to == pr) continue;
    dfs(to, v, d + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  g.resize(n + 1);
  vector<bool> used(n, 0);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v), g[v].push_back(u);
  }
  int ans2 = n - 1;
  for (int i = 1; i <= n; ++i) {
    if (used[i]) continue;
    if (g[i].size() == 1) {
      used[i] = 1;
      int v = g[i][0];
      for (int to : g[v]) {
        if (!used[to] && g[to].size() == 1) {
          ans2--, used[to] = 1;
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (g[i].size() == 1) {
      dfs(i, -1, 0);
      break;
    }
  }
  cout << ans1 << " " << ans2;
}

### Prompt

Please create a solution in CPP 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>
#pragma GCC optimize("O3")
using namespace std;
long long ppow(long long x, long long y, long long mod) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    y >>= 1;
    x = x * x % mod;
  }
  return res;
}
const int N = 1e5 + 5;
vector<vector<int>> g;
int ans1 = 1;
void dfs(int v, int pr, int d) {
  if (g[v].size() == 1 && d % 2) {
    ans1 = 3;
    return;
  }
  for (int to : g[v]) {
    if (to == pr) continue;
    dfs(to, v, d + 1);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  g.resize(n + 1);
  vector<bool> used(n, 0);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v), g[v].push_back(u);
  }
  int ans2 = n - 1;
  for (int i = 1; i <= n; ++i) {
    if (used[i]) continue;
    if (g[i].size() == 1) {
      used[i] = 1;
      int v = g[i][0];
      for (int to : g[v]) {
        if (!used[to] && g[to].size() == 1) {
          ans2--, used[to] = 1;
        }
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    if (g[i].size() == 1) {
      dfs(i, -1, 0);
      break;
    }
  }
  cout << ans1 << " " << ans2;
}

```