#include <bits/stdc++.h>
using namespace std;
long long mod = 1e9 + 7;
long long gcdll(long long a, long long b) {
  if (!b) return a;
  return gcdll(b, a % b);
}
void seive(int n) {
  vector<bool> is_prime(n + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (int j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
vector<int> constructZFunc(string s) {
  int n = s.size();
  vector<int> z(n);
  int l, r;
  l = r = 0;
  z[0] = 0;
  for (int i = 1; i <= n - 1; i++) {
    z[i] = 0;
    if (i <= r) {
      z[i] = min(r - i + 1, z[i - l]);
    }
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) {
      r = i + z[i] - 1;
      l = i;
    }
  }
  return z;
}
const int N = 1e5 + 5;
vector<vector<int>> g(N);
set<int> s;
void dfs(int v, int p, bool par, int &mx) {
  int cnt = 0;
  for (auto u : g[v]) {
    if (g[u].size() == 1) cnt++;
    if (u == p) continue;
    dfs(u, v, !par, mx);
  }
  if (g[v].size() == 1)
    s.insert(par);
  else if (cnt)
    mx -= (cnt - 1);
}
void solve() {
  int n, u, v, mx, mn;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  mx = n - 1;
  mn = 1;
  dfs(1, 1, 0, mx);
  if (s.size() > 1) mn = 3;
  cout << mn << " " << mx;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
long long mod = 1e9 + 7;
long long gcdll(long long a, long long b) {
  if (!b) return a;
  return gcdll(b, a % b);
}
void seive(int n) {
  vector<bool> is_prime(n + 1, true);
  is_prime[0] = is_prime[1] = false;
  for (int i = 2; i * i <= n; i++) {
    if (is_prime[i]) {
      for (int j = i * i; j <= n; j += i) {
        is_prime[j] = false;
      }
    }
  }
}
vector<int> constructZFunc(string s) {
  int n = s.size();
  vector<int> z(n);
  int l, r;
  l = r = 0;
  z[0] = 0;
  for (int i = 1; i <= n - 1; i++) {
    z[i] = 0;
    if (i <= r) {
      z[i] = min(r - i + 1, z[i - l]);
    }
    while (i + z[i] < n && s[z[i]] == s[i + z[i]]) ++z[i];
    if (i + z[i] - 1 > r) {
      r = i + z[i] - 1;
      l = i;
    }
  }
  return z;
}
const int N = 1e5 + 5;
vector<vector<int>> g(N);
set<int> s;
void dfs(int v, int p, bool par, int &mx) {
  int cnt = 0;
  for (auto u : g[v]) {
    if (g[u].size() == 1) cnt++;
    if (u == p) continue;
    dfs(u, v, !par, mx);
  }
  if (g[v].size() == 1)
    s.insert(par);
  else if (cnt)
    mx -= (cnt - 1);
}
void solve() {
  int n, u, v, mx, mn;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  mx = n - 1;
  mn = 1;
  dfs(1, 1, 0, mx);
  if (s.size() > 1) mn = 3;
  cout << mn << " " << mx;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```