#include <bits/stdc++.h>
using namespace std;
namespace RA {
int r(int p) { return 1ll * rand() * rand() % p; }
int r(int L, int R) { return r(R - L + 1) + L; }
}  // namespace RA
const int N = 1e5 + 5;
int n, dg[N], sz[N], dep[N], flag[2], f[N];
vector<int> g[N];
void dfs(int u, int p) {
  dep[u] = dep[p] + 1, sz[u] = 1;
  if (dg[u] == 1) flag[dep[u] & 1] = 1;
  for (int v : g[u])
    if (v != p) dfs(v, u), sz[u] += sz[v];
  int s0 = 0;
  for (int v : g[u])
    if (v != p) {
      if (sz[v] == 1)
        s0 = 1;
      else
        f[u] += f[v] + 1;
    }
  f[u] += s0;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b), g[b].push_back(a);
    dg[a]++, dg[b]++;
  }
  int root = max_element(dg + 1, dg + n + 1) - dg;
  dfs(root, 0);
  if (flag[0] & flag[1])
    printf("3 ");
  else
    printf("1 ");
  printf("%d\n", f[root]);
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
namespace RA {
int r(int p) { return 1ll * rand() * rand() % p; }
int r(int L, int R) { return r(R - L + 1) + L; }
}  // namespace RA
const int N = 1e5 + 5;
int n, dg[N], sz[N], dep[N], flag[2], f[N];
vector<int> g[N];
void dfs(int u, int p) {
  dep[u] = dep[p] + 1, sz[u] = 1;
  if (dg[u] == 1) flag[dep[u] & 1] = 1;
  for (int v : g[u])
    if (v != p) dfs(v, u), sz[u] += sz[v];
  int s0 = 0;
  for (int v : g[u])
    if (v != p) {
      if (sz[v] == 1)
        s0 = 1;
      else
        f[u] += f[v] + 1;
    }
  f[u] += s0;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n - 1); ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b), g[b].push_back(a);
    dg[a]++, dg[b]++;
  }
  int root = max_element(dg + 1, dg + n + 1) - dg;
  dfs(root, 0);
  if (flag[0] & flag[1])
    printf("3 ");
  else
    printf("1 ");
  printf("%d\n", f[root]);
  return 0;
}

```