#include <bits/stdc++.h>
using namespace std;
inline long long getnum() {
  char c = getchar();
  long long num, sign = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') sign = -1;
  for (num = 0; c >= '0' && c <= '9';) {
    c -= '0';
    num = num * 10 + c;
    c = getchar();
  }
  return num * sign;
}
vector<int> V[100004];
int X[100004][2];
void dfs(int v, int p) {
  if (V[v].size() == 1) {
    X[v][0] = 1;
  }
  for (int i : V[v]) {
    if (i == p) continue;
    dfs(i, v);
    if (X[i][0]) X[v][1] = 1;
    if (X[i][1]) X[v][0] = 1;
  }
}
int main() {
  int n = getnum();
  if (n == 2) {
    printf("1 1");
    return 0;
  }
  for (int i = 1; i < n; i++) {
    int x = getnum(), y = getnum();
    V[x].push_back(y);
    V[y].push_back(x);
  }
  int st = 0, cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (V[i].size() > 1) {
      st = i;
      cnt++;
      for (int x : V[i]) {
        if (V[x].size() == 1) {
          cnt++;
          break;
        }
      }
    }
  }
  cnt--;
  dfs(st, 0);
  if (X[st][0] && X[st][1]) {
    printf("3 %d", cnt);
  } else {
    printf("1 %d", cnt);
  }
}

### Prompt

Your challenge is to write 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>
using namespace std;
inline long long getnum() {
  char c = getchar();
  long long num, sign = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') sign = -1;
  for (num = 0; c >= '0' && c <= '9';) {
    c -= '0';
    num = num * 10 + c;
    c = getchar();
  }
  return num * sign;
}
vector<int> V[100004];
int X[100004][2];
void dfs(int v, int p) {
  if (V[v].size() == 1) {
    X[v][0] = 1;
  }
  for (int i : V[v]) {
    if (i == p) continue;
    dfs(i, v);
    if (X[i][0]) X[v][1] = 1;
    if (X[i][1]) X[v][0] = 1;
  }
}
int main() {
  int n = getnum();
  if (n == 2) {
    printf("1 1");
    return 0;
  }
  for (int i = 1; i < n; i++) {
    int x = getnum(), y = getnum();
    V[x].push_back(y);
    V[y].push_back(x);
  }
  int st = 0, cnt = 0;
  for (int i = 1; i <= n; i++) {
    if (V[i].size() > 1) {
      st = i;
      cnt++;
      for (int x : V[i]) {
        if (V[x].size() == 1) {
          cnt++;
          break;
        }
      }
    }
  }
  cnt--;
  dfs(st, 0);
  if (X[st][0] && X[st][1]) {
    printf("3 %d", cnt);
  } else {
    printf("1 %d", cnt);
  }
}

```