#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void swap(long long int &a, long long int &b) {
  long long int temp = a;
  a = b;
  b = temp;
}
long long int fast(long long int a, long long int m) {
  if (m == 1) return a % 998244353;
  if (m == 0) return 1;
  long long int temp = fast(a, m / 2);
  temp = (temp * temp) % 998244353;
  if (m % 2) return (temp * a) % 998244353;
  return temp;
}
void pre() {}
vector<vector<long long int> > adj;
void dfs(long long int v, long long int p, long long int &min,
         long long int &max, long long int &even, long long int &odd) {
  long long int even1 = 0, odd1 = 0, min1 = 1, max1 = 0;
  int flag = 0;
  for (auto u : adj[v]) {
    if (u != p) {
      if (adj[u].size() == 1) {
        odd1++;
        if (!flag) max1++;
        flag = 1;
      } else {
        min = 1, max = 0, odd = 0, even = 0;
        dfs(u, v, min, max, even, odd);
        odd1 += even;
        even1 += odd;
        max1 += (max + 1);
        if (min1 != 3) min1 = min;
      }
    }
  }
  max = max1;
  if (min1 == 1 && (even1 != 0 && odd1 != 0)) min1 = 3;
  min = min1;
  even = even1;
  odd = odd1;
}
void solve() {
  long long int n;
  cin >> n;
  adj.resize(n);
  for (long long int i = 0; i < (n - 1); ++i) {
    long long int a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  long long int root = -1;
  for (long long int i = 0; i < (n); ++i) {
    if (adj[i].size() >= 2) {
      root = i;
      break;
    }
  }
  long long int min = 1, max = 0, even = 0, odd = 0;
  dfs(root, root, min, max, even, odd);
  cout << min << " " << max;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) solve();
  return 0;
}

### Prompt

Generate a Cpp solution to 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 int gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
void swap(long long int &a, long long int &b) {
  long long int temp = a;
  a = b;
  b = temp;
}
long long int fast(long long int a, long long int m) {
  if (m == 1) return a % 998244353;
  if (m == 0) return 1;
  long long int temp = fast(a, m / 2);
  temp = (temp * temp) % 998244353;
  if (m % 2) return (temp * a) % 998244353;
  return temp;
}
void pre() {}
vector<vector<long long int> > adj;
void dfs(long long int v, long long int p, long long int &min,
         long long int &max, long long int &even, long long int &odd) {
  long long int even1 = 0, odd1 = 0, min1 = 1, max1 = 0;
  int flag = 0;
  for (auto u : adj[v]) {
    if (u != p) {
      if (adj[u].size() == 1) {
        odd1++;
        if (!flag) max1++;
        flag = 1;
      } else {
        min = 1, max = 0, odd = 0, even = 0;
        dfs(u, v, min, max, even, odd);
        odd1 += even;
        even1 += odd;
        max1 += (max + 1);
        if (min1 != 3) min1 = min;
      }
    }
  }
  max = max1;
  if (min1 == 1 && (even1 != 0 && odd1 != 0)) min1 = 3;
  min = min1;
  even = even1;
  odd = odd1;
}
void solve() {
  long long int n;
  cin >> n;
  adj.resize(n);
  for (long long int i = 0; i < (n - 1); ++i) {
    long long int a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  long long int root = -1;
  for (long long int i = 0; i < (n); ++i) {
    if (adj[i].size() >= 2) {
      root = i;
      break;
    }
  }
  long long int min = 1, max = 0, even = 0, odd = 0;
  dfs(root, root, min, max, even, odd);
  cout << min << " " << max;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) solve();
  return 0;
}

```