#include <bits/stdc++.h>
using namespace std;
long long int powmod(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
int nxt() {
  int x;
  cin >> x;
  return x;
}
vector<int> parent;
vector<vector<int> > adjList;
vector<pair<int, int> > leaves;
void dfs(int i, int d, int p) {
  for (int j = 0; j < (int)adjList[i].size(); j += 1)
    if (adjList[i][j] != p) {
      dfs(adjList[i][j], d + 1, i);
    }
  if (adjList[i].size() == 1) leaves.push_back(make_pair(d, adjList[i][0]));
}
int main() {
  int n = nxt();
  adjList.assign(n, vector<int>());
  for (int i = 0; i < n - 1; i += 1) {
    int a = nxt(), b = nxt();
    adjList[a - 1].push_back(b - 1);
    adjList[b - 1].push_back(a - 1);
  }
  dfs(0, 0, -1);
  bool even = true;
  bool parity = leaves[0].first % 2;
  for (int i = 0; i < (int)leaves.size(); i += 1)
    if (leaves[i].first % 2 != parity) {
      even = false;
      break;
    }
  int small = 3, big = n - 1;
  if (even) small = 1;
  map<int, int> mah;
  for (int i = 0; i < (int)leaves.size(); i += 1) mah[leaves[i].second]++;
  for (auto it = mah.begin(); it != mah.end(); it++) {
    big -= it->second - 1;
  }
  printf("%d %d\n", small, big);
}

### Prompt

Please create a solution in cpp 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 powmod(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long int gcd(long long int a, long long int b) {
  return b ? gcd(b, a % b) : a;
}
int nxt() {
  int x;
  cin >> x;
  return x;
}
vector<int> parent;
vector<vector<int> > adjList;
vector<pair<int, int> > leaves;
void dfs(int i, int d, int p) {
  for (int j = 0; j < (int)adjList[i].size(); j += 1)
    if (adjList[i][j] != p) {
      dfs(adjList[i][j], d + 1, i);
    }
  if (adjList[i].size() == 1) leaves.push_back(make_pair(d, adjList[i][0]));
}
int main() {
  int n = nxt();
  adjList.assign(n, vector<int>());
  for (int i = 0; i < n - 1; i += 1) {
    int a = nxt(), b = nxt();
    adjList[a - 1].push_back(b - 1);
    adjList[b - 1].push_back(a - 1);
  }
  dfs(0, 0, -1);
  bool even = true;
  bool parity = leaves[0].first % 2;
  for (int i = 0; i < (int)leaves.size(); i += 1)
    if (leaves[i].first % 2 != parity) {
      even = false;
      break;
    }
  int small = 3, big = n - 1;
  if (even) small = 1;
  map<int, int> mah;
  for (int i = 0; i < (int)leaves.size(); i += 1) mah[leaves[i].second]++;
  for (auto it = mah.begin(); it != mah.end(); it++) {
    big -= it->second - 1;
  }
  printf("%d %d\n", small, big);
}

```