#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.141592653589793238462643383279;
const long long MOD = 1e15;
const long long INF = 1e9;
const long double eps = 0.000000000000001;
template <class T>
inline istream& operator>>(istream& str, vector<T>& a) {
  for (auto& i : a) str >> i;
  return str;
}
void boot_up() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.precision(15);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
vector<vector<long long>> field(1e5 + 1);
vector<long long> st, st2;
vector<long long> used(1e5 + 1);
void dfs1(long long v, long long ptr) {
  if (field[v].size() == 1) {
    st.push_back(ptr);
  }
  used[v] = 1;
  for (auto i : field[v]) {
    if (!used[i]) {
      dfs1(i, ptr + 1);
    }
  }
}
void dfs2(long long v) {
  if (field[v].size() == 1) {
    st2.push_back(field[v][0]);
  }
  used[v] = 1;
  for (auto i : field[v]) {
    if (!used[i]) {
      dfs2(i);
    }
  }
}
signed main() {
  boot_up();
  long long t;
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    field[u].push_back(v);
    field[v].push_back(u);
  }
  long long ptr;
  for (long long i = 0; i <= 100000; i++) {
    if (field[i].size() == 1) {
      dfs1(i, 1);
      ptr = i;
      break;
    }
  }
  long long f_min = 1, f_max = n - 1;
  for (auto i : st) {
    if (i % 2 != st[0] % 2) {
      f_min = 3;
    }
  }
  cout << f_min << endl;
  for (long long i = 0; i <= n; i++) used[i] = 0;
  dfs2(ptr);
  set<long long> s;
  for (auto i : st2) {
    s.insert(i);
  }
  f_max -= (st2.size() - s.size());
  cout << f_max << endl;
  return 0;
}

### 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>
using namespace std;
const long double pi = 3.141592653589793238462643383279;
const long long MOD = 1e15;
const long long INF = 1e9;
const long double eps = 0.000000000000001;
template <class T>
inline istream& operator>>(istream& str, vector<T>& a) {
  for (auto& i : a) str >> i;
  return str;
}
void boot_up() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.setf(ios::fixed);
  cout.precision(15);
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
vector<vector<long long>> field(1e5 + 1);
vector<long long> st, st2;
vector<long long> used(1e5 + 1);
void dfs1(long long v, long long ptr) {
  if (field[v].size() == 1) {
    st.push_back(ptr);
  }
  used[v] = 1;
  for (auto i : field[v]) {
    if (!used[i]) {
      dfs1(i, ptr + 1);
    }
  }
}
void dfs2(long long v) {
  if (field[v].size() == 1) {
    st2.push_back(field[v][0]);
  }
  used[v] = 1;
  for (auto i : field[v]) {
    if (!used[i]) {
      dfs2(i);
    }
  }
}
signed main() {
  boot_up();
  long long t;
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    field[u].push_back(v);
    field[v].push_back(u);
  }
  long long ptr;
  for (long long i = 0; i <= 100000; i++) {
    if (field[i].size() == 1) {
      dfs1(i, 1);
      ptr = i;
      break;
    }
  }
  long long f_min = 1, f_max = n - 1;
  for (auto i : st) {
    if (i % 2 != st[0] % 2) {
      f_min = 3;
    }
  }
  cout << f_min << endl;
  for (long long i = 0; i <= n; i++) used[i] = 0;
  dfs2(ptr);
  set<long long> s;
  for (auto i : st2) {
    s.insert(i);
  }
  f_max -= (st2.size() - s.size());
  cout << f_max << endl;
  return 0;
}

```