#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int N;
vector<vector<int> > adj;
vector<int> is_leaf;
int ans_max;
void dfs2(int u, int p = -1) {
  int nchildren = 0;
  int nleaves = 0;
  for (int v : adj[u]) {
    if (v == p) continue;
    ++nchildren;
    dfs2(v, u);
    if (is_leaf[v]) ++nleaves;
  }
  if (nchildren == 0) {
    is_leaf[u] = true;
    return;
  }
  ans_max += nchildren - nleaves;
  if (nleaves >= 1) ans_max++;
}
int nodd = 0, neven = 0;
void dfs1(int u, int p = -1, int d = 0) {
  for (int v : adj[u]) {
    if (v == p) continue;
    dfs1(v, u, d + 1);
  }
  if (is_leaf[u]) {
    if (d & 1)
      ++nodd;
    else
      ++neven;
  }
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  adj = vector<vector<int> >(N);
  for (int j = 0, _n = (N - 1); j < _n; ++j) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int root = 0;
  for (int u = 0, _n = (N); u < _n; ++u)
    if (int((adj[u]).size()) > 1) {
      root = u;
      break;
    }
  is_leaf = vector<int>(N);
  dfs2(root);
  dfs1(root);
  int ans_min = nodd > 0 && neven > 0 ? 3 : 1;
  cout << ans_min << ' ' << ans_max << '\n';
  return 0;
}

### Prompt

Develop a solution in Cpp 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;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
int N;
vector<vector<int> > adj;
vector<int> is_leaf;
int ans_max;
void dfs2(int u, int p = -1) {
  int nchildren = 0;
  int nleaves = 0;
  for (int v : adj[u]) {
    if (v == p) continue;
    ++nchildren;
    dfs2(v, u);
    if (is_leaf[v]) ++nleaves;
  }
  if (nchildren == 0) {
    is_leaf[u] = true;
    return;
  }
  ans_max += nchildren - nleaves;
  if (nleaves >= 1) ans_max++;
}
int nodd = 0, neven = 0;
void dfs1(int u, int p = -1, int d = 0) {
  for (int v : adj[u]) {
    if (v == p) continue;
    dfs1(v, u, d + 1);
  }
  if (is_leaf[u]) {
    if (d & 1)
      ++nodd;
    else
      ++neven;
  }
}
int main(int argc, char *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int N;
  cin >> N;
  adj = vector<vector<int> >(N);
  for (int j = 0, _n = (N - 1); j < _n; ++j) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int root = 0;
  for (int u = 0, _n = (N); u < _n; ++u)
    if (int((adj[u]).size()) > 1) {
      root = u;
      break;
    }
  is_leaf = vector<int>(N);
  dfs2(root);
  dfs1(root);
  int ans_min = nodd > 0 && neven > 0 ? 3 : 1;
  cout << ans_min << ' ' << ans_max << '\n';
  return 0;
}

```