#include <bits/stdc++.h>
const int maxn = 100000 + 10;
std::vector<int> G[maxn], leaves;
int dep[maxn];
bool isleaf[maxn];
void dfs(int u, int p = 0, int d = 0) {
  dep[u] = d;
  for (auto v : G[u])
    if (v != p) {
      dfs(v, u, d + 1);
    }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (G[i].size() == 1) {
      leaves.push_back(i);
      isleaf[i] = true;
    }
  int min = 1;
  int odd = -1;
  dfs(1);
  for (int leaf : leaves) {
    if (odd == -1)
      odd = dep[leaf] & 1;
    else if ((dep[leaf] & 1) ^ odd) {
      min = 3;
      break;
    }
  }
  int max = n - 1 - leaves.size();
  for (int i = 1; i <= n; i++) {
    for (int j : G[i])
      if (isleaf[j]) {
        max++;
        break;
      }
  }
  printf("%d %d\n", min, max);
  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>
const int maxn = 100000 + 10;
std::vector<int> G[maxn], leaves;
int dep[maxn];
bool isleaf[maxn];
void dfs(int u, int p = 0, int d = 0) {
  dep[u] = d;
  for (auto v : G[u])
    if (v != p) {
      dfs(v, u, d + 1);
    }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i <= n; i++)
    if (G[i].size() == 1) {
      leaves.push_back(i);
      isleaf[i] = true;
    }
  int min = 1;
  int odd = -1;
  dfs(1);
  for (int leaf : leaves) {
    if (odd == -1)
      odd = dep[leaf] & 1;
    else if ((dep[leaf] & 1) ^ odd) {
      min = 3;
      break;
    }
  }
  int max = n - 1 - leaves.size();
  for (int i = 1; i <= n; i++) {
    for (int j : G[i])
      if (isleaf[j]) {
        max++;
        break;
      }
  }
  printf("%d %d\n", min, max);
  return 0;
}

```