#include <bits/stdc++.h>
using namespace std;
int max_of_num;
int n, test_Q, point_num[100005], f_delp[100005][2];
vector<int> vc[100005];
bool flag = false;
void search1(int u, int fa) {
  for (int i = 0; i < vc[u].size(); i++) {
    int v = vc[u][i];
    if (v != fa) {
      if (point_num[v] == 1) {
        if (f_delp[u][0]) {
          flag = true;
        }
        f_delp[u][1] = true;
      } else {
        search1(v, u);
        if (f_delp[v][1] && f_delp[u][1]) {
          flag = true;
        }
        if (f_delp[v][0] && f_delp[u][0]) {
          flag = true;
        }
        f_delp[u][0] = f_delp[u][0] | f_delp[v][1];
        f_delp[u][1] = f_delp[u][1] | f_delp[v][0];
      }
    }
  }
}
void connect_add(int u = 0, int v = 0) {
  scanf("%d%d", &u, &v);
  point_num[u]++;
  point_num[v]++;
  vc[u].push_back(v);
  vc[v].push_back(u);
}
void search2(int u, int fa) {
  int qaq = 0;
  for (int i = 0; i < vc[u].size(); i++) {
    int v = vc[u][i];
    if (v != fa) {
      if (point_num[v] == 1)
        qaq++;
      else
        search2(v, u);
    }
  }
  if (qaq) max_of_num = max_of_num - (qaq - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    connect_add();
  }
  test_Q = 0;
  for (int i = 1; i <= n; ++i) {
    if (point_num[i] > 1) {
      test_Q = i;
    }
  }
  search1(test_Q, 0);
  if (flag) {
    printf("3 ");
  } else {
    printf("1 ");
  }
  max_of_num = n - 1;
  search2(test_Q, 0);
  printf("%d\n", max_of_num);
  return 0;
}

### 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>
using namespace std;
int max_of_num;
int n, test_Q, point_num[100005], f_delp[100005][2];
vector<int> vc[100005];
bool flag = false;
void search1(int u, int fa) {
  for (int i = 0; i < vc[u].size(); i++) {
    int v = vc[u][i];
    if (v != fa) {
      if (point_num[v] == 1) {
        if (f_delp[u][0]) {
          flag = true;
        }
        f_delp[u][1] = true;
      } else {
        search1(v, u);
        if (f_delp[v][1] && f_delp[u][1]) {
          flag = true;
        }
        if (f_delp[v][0] && f_delp[u][0]) {
          flag = true;
        }
        f_delp[u][0] = f_delp[u][0] | f_delp[v][1];
        f_delp[u][1] = f_delp[u][1] | f_delp[v][0];
      }
    }
  }
}
void connect_add(int u = 0, int v = 0) {
  scanf("%d%d", &u, &v);
  point_num[u]++;
  point_num[v]++;
  vc[u].push_back(v);
  vc[v].push_back(u);
}
void search2(int u, int fa) {
  int qaq = 0;
  for (int i = 0; i < vc[u].size(); i++) {
    int v = vc[u][i];
    if (v != fa) {
      if (point_num[v] == 1)
        qaq++;
      else
        search2(v, u);
    }
  }
  if (qaq) max_of_num = max_of_num - (qaq - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 2; i <= n; i++) {
    connect_add();
  }
  test_Q = 0;
  for (int i = 1; i <= n; ++i) {
    if (point_num[i] > 1) {
      test_Q = i;
    }
  }
  search1(test_Q, 0);
  if (flag) {
    printf("3 ");
  } else {
    printf("1 ");
  }
  max_of_num = n - 1;
  search2(test_Q, 0);
  printf("%d\n", max_of_num);
  return 0;
}

```