#include <bits/stdc++.h>
using namespace std;
const long long N = 2e6 + 5;
const long long INF = 1e18L;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
long long vis[N];
long long dp[N];
map<long long, long long> mp, mp2;
vector<long long> v[N];
map<long long, long long> NoN_leaf;
vector<long long> leaf;
void dfs(long long s) {
  vis[s] = 1;
  for (auto i : v[s]) {
    if (!vis[i]) {
      dp[i] = 1 + dp[s];
      dfs(i);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long T = 1;
  while (T--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < n - 1; i++) {
      long long x, y;
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
      mp[x]++;
      mp[y]++;
    }
    for (long long i = 1; i <= n; i++) {
      if (mp[i] == 1) leaf.push_back(i);
    }
    for (auto i : leaf) {
      for (auto j : v[i]) NoN_leaf[j]++;
    }
    dfs(leaf[0]);
    long long o, e;
    o = e = 0;
    for (long long i = 0; i < leaf.size(); i++) {
      long long d = dp[leaf[i]] % 2;
      if (d & 1)
        o++;
      else
        e++;
    }
    if (o)
      cout << "3 ";
    else
      cout << "1 ";
    cout << n - 1 - leaf.size() + NoN_leaf.size();
  }
  cerr << "Time Taken : " << 1.0 * clock() / CLOCKS_PER_SEC << " sec \n";
  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 long N = 2e6 + 5;
const long long INF = 1e18L;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cerr << name << ": " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << ": " << arg1 << " |";
  __f(comma + 1, args...);
}
long long vis[N];
long long dp[N];
map<long long, long long> mp, mp2;
vector<long long> v[N];
map<long long, long long> NoN_leaf;
vector<long long> leaf;
void dfs(long long s) {
  vis[s] = 1;
  for (auto i : v[s]) {
    if (!vis[i]) {
      dp[i] = 1 + dp[s];
      dfs(i);
    }
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long T = 1;
  while (T--) {
    long long n;
    cin >> n;
    for (long long i = 0; i < n - 1; i++) {
      long long x, y;
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
      mp[x]++;
      mp[y]++;
    }
    for (long long i = 1; i <= n; i++) {
      if (mp[i] == 1) leaf.push_back(i);
    }
    for (auto i : leaf) {
      for (auto j : v[i]) NoN_leaf[j]++;
    }
    dfs(leaf[0]);
    long long o, e;
    o = e = 0;
    for (long long i = 0; i < leaf.size(); i++) {
      long long d = dp[leaf[i]] % 2;
      if (d & 1)
        o++;
      else
        e++;
    }
    if (o)
      cout << "3 ";
    else
      cout << "1 ";
    cout << n - 1 - leaf.size() + NoN_leaf.size();
  }
  cerr << "Time Taken : " << 1.0 * clock() / CLOCKS_PER_SEC << " sec \n";
  return 0;
}

```