#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
const int N = (int)5e5 + 100;
const int M = (int)2e6 + 100;
const int inf = (int)1e9 + 100;
vector<int> v[N];
pair<bool, bool> dfs(int x, int pr = -1) {
  pair<bool, bool> o = {0, 0};
  if (v[x].size() == 1) o.first = 1;
  for (auto y : v[x]) {
    if (y == pr) continue;
    pair<bool, bool> c = dfs(y, x);
    o.first = max(o.first, c.second);
    o.second = max(o.second, c.first);
  }
  return o;
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, i, a, b;
  cin >> n;
  for ((i) = 0; (i) < (n - 1); (i)++) {
    cin >> a >> b;
    a--;
    b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  vector<int> g;
  for ((i) = 0; (i) < (n); (i)++) {
    if (v[i].size() == 1) {
      g.push_back(v[i][0]);
    }
  }
  a = g.size();
  sort((g).begin(), (g).end()),
      g.resize(unique((g).begin(), (g).end()) - g.begin());
  b = g.size();
  pair<bool, bool> cc = dfs(0);
  if (cc.first && cc.second) {
    cout << 3;
  } else {
    cout << 1;
  }
  cout << ' ' << n - 1 - (a - b);
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
const int N = (int)5e5 + 100;
const int M = (int)2e6 + 100;
const int inf = (int)1e9 + 100;
vector<int> v[N];
pair<bool, bool> dfs(int x, int pr = -1) {
  pair<bool, bool> o = {0, 0};
  if (v[x].size() == 1) o.first = 1;
  for (auto y : v[x]) {
    if (y == pr) continue;
    pair<bool, bool> c = dfs(y, x);
    o.first = max(o.first, c.second);
    o.second = max(o.second, c.first);
  }
  return o;
}
int main() {
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  srand(time(0));
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, i, a, b;
  cin >> n;
  for ((i) = 0; (i) < (n - 1); (i)++) {
    cin >> a >> b;
    a--;
    b--;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  vector<int> g;
  for ((i) = 0; (i) < (n); (i)++) {
    if (v[i].size() == 1) {
      g.push_back(v[i][0]);
    }
  }
  a = g.size();
  sort((g).begin(), (g).end()),
      g.resize(unique((g).begin(), (g).end()) - g.begin());
  b = g.size();
  pair<bool, bool> cc = dfs(0);
  if (cc.first && cc.second) {
    cout << 3;
  } else {
    cout << 1;
  }
  cout << ' ' << n - 1 - (a - b);
}

```