#include <bits/stdc++.h>
using namespace std;
void pv(vector<long long> v) {
  for (long long i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "" << endl;
}
void pa(long long a[], long long n, long long p) {
  for (long long i = p; i < n + p; ++i) {
    cout << a[i] << " ";
  }
  cout << "" << endl;
}
const double pi = 3.141592653589793238;
const long long inf = (long long)(10e18);
const long long mod = 1000000007;
const long long M = 2000;
long long modexp(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1LL) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long invmod(long long n) { return modexp(n, mod - 2); }
std::vector<long long> v[100005];
long long check[2], dif;
void dfs(long long node, long long p, long long h) {
  long long u, i;
  if (v[node].size() == 1) check[h % 2]++;
  for (i = 0; i < v[node].size(); i++) {
    u = v[node][i];
    if (u != p) {
      dfs(u, node, h + 1);
    }
  }
}
void dfs1(long long node, long long p, long long path, long long ww) {
  long long u, i, flag = 0;
  if (v[node].size() == 1) {
    if (ww == 1) path++;
    if (path != 1) dif += path;
  } else if (v[node].size() > 2) {
    path++;
    if (path != 1) dif += path;
    path = 0;
  } else {
    path++;
  }
  for (i = 0; i < v[node].size(); i++) {
    u = v[node][i];
    if (u != p) {
      dfs1(u, node, path, ww);
    }
  }
}
long long ans = 0;
long long dfs2(long long node, long long p) {
  long long u, i, pp = 0;
  if (v[node].size() == 1) {
    return 1;
  }
  for (i = 0; i < v[node].size(); i++) {
    u = v[node][i];
    if (u != p) {
      pp += dfs2(u, node);
    }
  }
  ans += max(0LL, pp - 1LL);
  return 0;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = (long long)1; i <= (long long)n - 1; i += (long long)1) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  long long root = -1;
  for (long long i = (long long)1; i <= (long long)n; i += (long long)1)
    if (v[i].size() > 2) root = i;
  if (root == -1) {
    if ((n - 1) % 2 == 0) {
      if (n == 3)
        cout << 1 << " " << 1 << endl;
      else
        cout << 1 << " " << n - 1 << endl;
    } else {
      cout << 3 << " " << n - 1 << endl;
    }
    return;
  }
  dfs(root, -1, 0);
  if (check[0] == 0 || check[1] == 0) {
    dfs2(root, -1);
    cout << 1 << " " << n - 1 - ans;
  } else {
    dfs2(root, -1);
    cout << 3 << " " << n - 1 - ans;
  }
}
int32_t main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
}

### Prompt

In Cpp, your task is to solve the following problem:

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 pv(vector<long long> v) {
  for (long long i = 0; i < v.size(); ++i) {
    cout << v[i] << " ";
  }
  cout << "" << endl;
}
void pa(long long a[], long long n, long long p) {
  for (long long i = p; i < n + p; ++i) {
    cout << a[i] << " ";
  }
  cout << "" << endl;
}
const double pi = 3.141592653589793238;
const long long inf = (long long)(10e18);
const long long mod = 1000000007;
const long long M = 2000;
long long modexp(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1LL) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long invmod(long long n) { return modexp(n, mod - 2); }
std::vector<long long> v[100005];
long long check[2], dif;
void dfs(long long node, long long p, long long h) {
  long long u, i;
  if (v[node].size() == 1) check[h % 2]++;
  for (i = 0; i < v[node].size(); i++) {
    u = v[node][i];
    if (u != p) {
      dfs(u, node, h + 1);
    }
  }
}
void dfs1(long long node, long long p, long long path, long long ww) {
  long long u, i, flag = 0;
  if (v[node].size() == 1) {
    if (ww == 1) path++;
    if (path != 1) dif += path;
  } else if (v[node].size() > 2) {
    path++;
    if (path != 1) dif += path;
    path = 0;
  } else {
    path++;
  }
  for (i = 0; i < v[node].size(); i++) {
    u = v[node][i];
    if (u != p) {
      dfs1(u, node, path, ww);
    }
  }
}
long long ans = 0;
long long dfs2(long long node, long long p) {
  long long u, i, pp = 0;
  if (v[node].size() == 1) {
    return 1;
  }
  for (i = 0; i < v[node].size(); i++) {
    u = v[node][i];
    if (u != p) {
      pp += dfs2(u, node);
    }
  }
  ans += max(0LL, pp - 1LL);
  return 0;
}
void solve() {
  long long n;
  cin >> n;
  for (long long i = (long long)1; i <= (long long)n - 1; i += (long long)1) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  long long root = -1;
  for (long long i = (long long)1; i <= (long long)n; i += (long long)1)
    if (v[i].size() > 2) root = i;
  if (root == -1) {
    if ((n - 1) % 2 == 0) {
      if (n == 3)
        cout << 1 << " " << 1 << endl;
      else
        cout << 1 << " " << n - 1 << endl;
    } else {
      cout << 3 << " " << n - 1 << endl;
    }
    return;
  }
  dfs(root, -1, 0);
  if (check[0] == 0 || check[1] == 0) {
    dfs2(root, -1);
    cout << 1 << " " << n - 1 - ans;
  } else {
    dfs2(root, -1);
    cout << 3 << " " << n - 1 - ans;
  }
}
int32_t main() {
  ios ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) solve();
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
}

```