#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
mt19937_64 rng(73982);
vector<vector<int> > adjLis;
int prefxor;
bool trav(int x, int p = -1) {
  if (adjLis[x].size() == 1 && prefxor != 0) {
    return false;
  }
  for (int y : adjLis[x]) {
    if (y == p) continue;
    prefxor ^= 1;
    if (!trav(y, x)) return false;
    prefxor ^= 1;
  }
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  adjLis.resize(n);
  int line = true;
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    adjLis[a].push_back(b);
    adjLis[b].push_back(a);
    if (adjLis[a].size() > 2) line = false;
    if (adjLis[b].size() > 2) line = false;
  }
  for (int i = 0; i < n; i++) {
    if (adjLis[i].size() == 1) {
      prefxor = 0;
      if (trav(i)) {
        printf("1 ");
      } else {
        printf("3 ");
      }
      break;
    }
  }
  map<pair<int, int>, long long> values;
  for (int i = 0; i < n; i++) {
    if (adjLis[i].size() != 1) {
      long long val = rng();
      for (int j : adjLis[i]) {
        if (i < j) {
          values[make_pair(i, j)] ^= val;
        } else {
          values[make_pair(j, i)] ^= val;
        }
      }
    }
  }
  set<long long> ans;
  for (auto i : values) {
    ans.insert(i.second);
  }
  printf("%lu\n", ans.size());
}

### 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;
const int mod = 1000000007;
mt19937_64 rng(73982);
vector<vector<int> > adjLis;
int prefxor;
bool trav(int x, int p = -1) {
  if (adjLis[x].size() == 1 && prefxor != 0) {
    return false;
  }
  for (int y : adjLis[x]) {
    if (y == p) continue;
    prefxor ^= 1;
    if (!trav(y, x)) return false;
    prefxor ^= 1;
  }
  return true;
}
int main() {
  int n;
  scanf("%d", &n);
  adjLis.resize(n);
  int line = true;
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    a--;
    b--;
    adjLis[a].push_back(b);
    adjLis[b].push_back(a);
    if (adjLis[a].size() > 2) line = false;
    if (adjLis[b].size() > 2) line = false;
  }
  for (int i = 0; i < n; i++) {
    if (adjLis[i].size() == 1) {
      prefxor = 0;
      if (trav(i)) {
        printf("1 ");
      } else {
        printf("3 ");
      }
      break;
    }
  }
  map<pair<int, int>, long long> values;
  for (int i = 0; i < n; i++) {
    if (adjLis[i].size() != 1) {
      long long val = rng();
      for (int j : adjLis[i]) {
        if (i < j) {
          values[make_pair(i, j)] ^= val;
        } else {
          values[make_pair(j, i)] ^= val;
        }
      }
    }
  }
  set<long long> ans;
  for (auto i : values) {
    ans.insert(i.second);
  }
  printf("%lu\n", ans.size());
}

```