#include <bits/stdc++.h>
using namespace std;
using ii = pair<int, int>;
using vi = vector<int>;
using vii = vector<ii>;
using vvi = vector<vi>;
using ll = long long int;
const int MOD = 1000 * 1000 * 1000 + 7;
int dfsmin(int nact, int nant, vvi const& listaAdy, bool& distImp) {
  if (nant != -1 && listaAdy[nact].size() == 1)
    return 1;
  else if (listaAdy[nact].size() > 1) {
    vi dists;
    for (int ady : listaAdy[nact]) {
      if (ady != nant) {
        dists.push_back(dfsmin(ady, nact, listaAdy, distImp));
      }
    }
    for (int i = 1; i < dists.size(); ++i) {
      if (dists[0] % 2 != dists[i] % 2) distImp = true;
    }
    return (dists[0] + 1) % 2;
  } else {
    if (dfsmin(listaAdy[nact][0], nact, listaAdy, distImp) == 1) distImp = true;
    return 0;
  }
}
int dfsmax(int nact, int nant, vvi const& listaAdy) {
  int sol = 0;
  bool leafs = false;
  for (int ady : listaAdy[nact]) {
    if (ady != nant) {
      if (listaAdy[ady].size() == 1)
        leafs = true;
      else
        sol += dfsmax(ady, nact, listaAdy) + 1;
    }
  }
  if (leafs) sol += 1;
  return sol;
}
int main() {
  int n;
  cin >> n;
  vvi listaAdy(n);
  for (int i = 0; i < n - 1; ++i) {
    int ini, fin;
    cin >> ini >> fin;
    --ini;
    --fin;
    listaAdy[ini].push_back(fin);
    listaAdy[fin].push_back(ini);
  }
  int rootleaf = -1;
  int rootnotleaf = -1;
  for (int i = 0; i < n; ++i) {
    if (listaAdy[i].size() == 1)
      rootleaf = i;
    else
      rootnotleaf = i;
    if (rootnotleaf != -1 && rootleaf != -1) break;
  }
  bool distImp = false;
  dfsmin(rootleaf, -1, listaAdy, distImp);
  if (distImp)
    cout << 3;
  else
    cout << 1;
  cout << ' ' << dfsmax(rootnotleaf, -1, listaAdy) << '\n';
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
using ii = pair<int, int>;
using vi = vector<int>;
using vii = vector<ii>;
using vvi = vector<vi>;
using ll = long long int;
const int MOD = 1000 * 1000 * 1000 + 7;
int dfsmin(int nact, int nant, vvi const& listaAdy, bool& distImp) {
  if (nant != -1 && listaAdy[nact].size() == 1)
    return 1;
  else if (listaAdy[nact].size() > 1) {
    vi dists;
    for (int ady : listaAdy[nact]) {
      if (ady != nant) {
        dists.push_back(dfsmin(ady, nact, listaAdy, distImp));
      }
    }
    for (int i = 1; i < dists.size(); ++i) {
      if (dists[0] % 2 != dists[i] % 2) distImp = true;
    }
    return (dists[0] + 1) % 2;
  } else {
    if (dfsmin(listaAdy[nact][0], nact, listaAdy, distImp) == 1) distImp = true;
    return 0;
  }
}
int dfsmax(int nact, int nant, vvi const& listaAdy) {
  int sol = 0;
  bool leafs = false;
  for (int ady : listaAdy[nact]) {
    if (ady != nant) {
      if (listaAdy[ady].size() == 1)
        leafs = true;
      else
        sol += dfsmax(ady, nact, listaAdy) + 1;
    }
  }
  if (leafs) sol += 1;
  return sol;
}
int main() {
  int n;
  cin >> n;
  vvi listaAdy(n);
  for (int i = 0; i < n - 1; ++i) {
    int ini, fin;
    cin >> ini >> fin;
    --ini;
    --fin;
    listaAdy[ini].push_back(fin);
    listaAdy[fin].push_back(ini);
  }
  int rootleaf = -1;
  int rootnotleaf = -1;
  for (int i = 0; i < n; ++i) {
    if (listaAdy[i].size() == 1)
      rootleaf = i;
    else
      rootnotleaf = i;
    if (rootnotleaf != -1 && rootleaf != -1) break;
  }
  bool distImp = false;
  dfsmin(rootleaf, -1, listaAdy, distImp);
  if (distImp)
    cout << 3;
  else
    cout << 1;
  cout << ' ' << dfsmax(rootnotleaf, -1, listaAdy) << '\n';
  return 0;
}

```