#include <bits/stdc++.h>
using namespace std;
template <typename T>
void __read(T &a) {
  cin >> a;
}
template <typename T, typename... Args>
void __read(T &a, Args &...args) {
  cin >> a;
  __read(args...);
}
constexpr long long M7 = 1000000007ll;
constexpr long long M9 = 1000000009ll;
constexpr long long MFFT = 998244353ll;
template <class T>
void outv(T &a) {
  for (auto &x : a) cout << x << ' ';
}
static mt19937 rnd(static_cast<unsigned>(
    chrono::steady_clock::now().time_since_epoch().count()));
auto __fast_io__ = (ios_base::sync_with_stdio(false), cin.tie(nullptr));
int32_t main() {
  long long n;
  __read(n);
  vector<vector<long long>> g(n);
  for (long long i = 1; i < n; ++i) {
    long long v, u;
    __read(v, u);
    v--, u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  vector<long long> sz(n, 1), d(n);
  long long mx_ans = n - 1;
  function<void(long long, long long)> dfs = [&](long long v, long long p) {
    long long cnt1 = 0;
    for (long long u : g[v]) {
      if (u != p) {
        d[u] = d[v] + 1;
        dfs(u, v);
        sz[v] += sz[u];
        if (sz[u] == 1) {
          cnt1++;
        }
      }
    }
    mx_ans -= max(0ll, cnt1 - 1);
  };
  long long v = 0;
  while (g[v].size() == 1) {
    v++;
  }
  dfs(v, -1);
  long long cnt[] = {0, 0};
  for (long long v = 0; v < n; ++v) {
    if (g[v].size() == 1) {
      cnt[d[v] & 1]++;
    }
  }
  cout << (cnt[0] > 0 && cnt[1] > 0 ? 3 : 1) << ' ' << mx_ans;
  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;
template <typename T>
void __read(T &a) {
  cin >> a;
}
template <typename T, typename... Args>
void __read(T &a, Args &...args) {
  cin >> a;
  __read(args...);
}
constexpr long long M7 = 1000000007ll;
constexpr long long M9 = 1000000009ll;
constexpr long long MFFT = 998244353ll;
template <class T>
void outv(T &a) {
  for (auto &x : a) cout << x << ' ';
}
static mt19937 rnd(static_cast<unsigned>(
    chrono::steady_clock::now().time_since_epoch().count()));
auto __fast_io__ = (ios_base::sync_with_stdio(false), cin.tie(nullptr));
int32_t main() {
  long long n;
  __read(n);
  vector<vector<long long>> g(n);
  for (long long i = 1; i < n; ++i) {
    long long v, u;
    __read(v, u);
    v--, u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  vector<long long> sz(n, 1), d(n);
  long long mx_ans = n - 1;
  function<void(long long, long long)> dfs = [&](long long v, long long p) {
    long long cnt1 = 0;
    for (long long u : g[v]) {
      if (u != p) {
        d[u] = d[v] + 1;
        dfs(u, v);
        sz[v] += sz[u];
        if (sz[u] == 1) {
          cnt1++;
        }
      }
    }
    mx_ans -= max(0ll, cnt1 - 1);
  };
  long long v = 0;
  while (g[v].size() == 1) {
    v++;
  }
  dfs(v, -1);
  long long cnt[] = {0, 0};
  for (long long v = 0; v < n; ++v) {
    if (g[v].size() == 1) {
      cnt[d[v] & 1]++;
    }
  }
  cout << (cnt[0] > 0 && cnt[1] > 0 ? 3 : 1) << ' ' << mx_ans;
  return 0;
}

```