#include <bits/stdc++.h>
const int N = 100100, M = N << 1;
int ft[N], nt[M], to[M], n, r, t = -1, s, tl;
bool dep[N], fl = 0;
inline void add(int x, int y) { to[++tl] = y, nt[tl] = ft[x], ft[x] = tl; }
void dfs(int x, int fa) {
  int c = 0;
  dep[x] = !dep[fa];
  for (int i = ft[x]; i; i = nt[i])
    if (!nt[ft[to[i]]]) {
      if (t == -1)
        t = dep[x];
      else if (dep[x] != t)
        fl = 1;
      c++;
    } else if (to[i] != fa)
      dfs(to[i], x);
  if (c) s -= c - 1;
}
int main() {
  register int i, t1, t2;
  scanf("%d", &n);
  s = n - 1;
  for (i = 1; i < n; i++) scanf("%d%d", &t1, &t2), add(t1, t2), add(t2, t1);
  if (!nt[ft[1]])
    r = to[ft[1]], ft[1] = 0;
  else
    r = 1;
  dfs(r, 0);
  if (fl)
    putchar(51);
  else
    putchar(49);
  printf(" %d", s);
}

### 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>
const int N = 100100, M = N << 1;
int ft[N], nt[M], to[M], n, r, t = -1, s, tl;
bool dep[N], fl = 0;
inline void add(int x, int y) { to[++tl] = y, nt[tl] = ft[x], ft[x] = tl; }
void dfs(int x, int fa) {
  int c = 0;
  dep[x] = !dep[fa];
  for (int i = ft[x]; i; i = nt[i])
    if (!nt[ft[to[i]]]) {
      if (t == -1)
        t = dep[x];
      else if (dep[x] != t)
        fl = 1;
      c++;
    } else if (to[i] != fa)
      dfs(to[i], x);
  if (c) s -= c - 1;
}
int main() {
  register int i, t1, t2;
  scanf("%d", &n);
  s = n - 1;
  for (i = 1; i < n; i++) scanf("%d%d", &t1, &t2), add(t1, t2), add(t2, t1);
  if (!nt[ft[1]])
    r = to[ft[1]], ft[1] = 0;
  else
    r = 1;
  dfs(r, 0);
  if (fl)
    putchar(51);
  else
    putchar(49);
  printf(" %d", s);
}

```