#include <bits/stdc++.h>
using namespace std;
vector<int> e[100010];
int du[100010], dep[100010], sub[100010], maxn = 0;
template <typename _T>
void rd(_T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s - '0'), s = getchar();
  x *= f;
}
void search(int x, int fa) {
  int lef = 0;
  for (auto y : e[x]) {
    if (y == fa) continue;
    dep[y] = dep[x] + 1;
    search(y, x);
    if (du[y] == 1) lef++;
    sub[x] = max(sub[y] + 1, sub[x]);
  }
  if (lef > 1) maxn += lef - 1;
  return;
}
int main() {
  int n;
  rd(n);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    rd(x), rd(y);
    e[x].push_back(y), e[y].push_back(x);
    du[x]++, du[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (du[i] > 1) {
      dep[i] = 1;
      search(i, -1);
      break;
    }
  int od = 0, ev = 0;
  for (int i = 1; i <= n; i++) {
    if (du[i] == 1 && (dep[i] & 1)) od = 1;
    if (du[i] == 1 && (dep[i] & 1) == 0) ev = 1;
  }
  int minn = 1, maxx = 0;
  if (od && ev) minn = 3;
  maxx = n - 1 - maxn;
  printf("%d %d\n", minn, maxx);
  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;
vector<int> e[100010];
int du[100010], dep[100010], sub[100010], maxn = 0;
template <typename _T>
void rd(_T &x) {
  int f = 1;
  x = 0;
  char s = getchar();
  while (s > '9' || s < '0') {
    if (s == '-') f = -1;
    s = getchar();
  }
  while (s >= '0' && s <= '9')
    x = (x << 3) + (x << 1) + (s - '0'), s = getchar();
  x *= f;
}
void search(int x, int fa) {
  int lef = 0;
  for (auto y : e[x]) {
    if (y == fa) continue;
    dep[y] = dep[x] + 1;
    search(y, x);
    if (du[y] == 1) lef++;
    sub[x] = max(sub[y] + 1, sub[x]);
  }
  if (lef > 1) maxn += lef - 1;
  return;
}
int main() {
  int n;
  rd(n);
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    rd(x), rd(y);
    e[x].push_back(y), e[y].push_back(x);
    du[x]++, du[y]++;
  }
  for (int i = 1; i <= n; i++)
    if (du[i] > 1) {
      dep[i] = 1;
      search(i, -1);
      break;
    }
  int od = 0, ev = 0;
  for (int i = 1; i <= n; i++) {
    if (du[i] == 1 && (dep[i] & 1)) od = 1;
    if (du[i] == 1 && (dep[i] & 1) == 0) ev = 1;
  }
  int minn = 1, maxx = 0;
  if (od && ev) minn = 3;
  maxx = n - 1 - maxn;
  printf("%d %d\n", minn, maxx);
  return 0;
}

```