#include <bits/stdc++.h>
using namespace std;
const int M = 1e5 + 10;
int res, cnt, head[M], in[M], dep[2];
struct node {
  int next, to;
} edge[M * 2];
void add_edge(int u, int v) {
  edge[++cnt].next = head[u];
  edge[cnt].to = v;
  head[u] = cnt;
}
void dfs(int u, int fa, int h) {
  if (in[u] == 1) dep[h % 2] = 1;
  int ans = 0;
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u, h + 1);
    if (in[v] == 1) ans++;
  }
  if (ans > 1) res += ans - 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
    in[u]++;
    in[v]++;
  }
  int root;
  for (int i = 1; i <= n; i++) {
    if (in[i] != 1) root = i;
  }
  dfs(root, 0, 1);
  if (dep[0] && dep[1])
    printf("3 ");
  else
    printf("1 ");
  printf("%d\n", (n - 1) - res);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int M = 1e5 + 10;
int res, cnt, head[M], in[M], dep[2];
struct node {
  int next, to;
} edge[M * 2];
void add_edge(int u, int v) {
  edge[++cnt].next = head[u];
  edge[cnt].to = v;
  head[u] = cnt;
}
void dfs(int u, int fa, int h) {
  if (in[u] == 1) dep[h % 2] = 1;
  int ans = 0;
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == fa) continue;
    dfs(v, u, h + 1);
    if (in[v] == 1) ans++;
  }
  if (ans > 1) res += ans - 1;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
    add_edge(v, u);
    in[u]++;
    in[v]++;
  }
  int root;
  for (int i = 1; i <= n; i++) {
    if (in[i] != 1) root = i;
  }
  dfs(root, 0, 1);
  if (dep[0] && dep[1])
    printf("3 ");
  else
    printf("1 ");
  printf("%d\n", (n - 1) - res);
  return 0;
}

```