#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 10;
typedef struct a {
  int to, next;
} per;
per si[maxn * 4];
int ctt, n, chu[maxn], head[maxn], ji, ou, dep[maxn], vis[maxn], siz[maxn], cnt;
void add(int u, int v) {
  si[++ctt].next = head[u];
  si[ctt].to = v;
  head[u] = ctt;
}
int dfs(int now, int fa) {
  dep[now] = dep[fa] + 1;
  int sum = 0;
  for (int i = head[now]; i != -1; i = si[i].next) {
    int to = si[i].to;
    if (to != fa) {
      sum += dfs(to, now);
    }
  }
  if (sum > 1) cnt += sum - 1;
  if (si[head[now]].to == fa && si[head[now]].next == -1) {
    if (dep[now] & 1)
      ji = 1;
    else
      ou = 1;
    return 1;
  }
  return 0;
}
int main() {
  cin >> n;
  int u, v;
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    add(u, v);
    add(v, u);
    chu[u]++;
    chu[v]++;
  }
  int root;
  for (int i = 1; i <= n; ++i)
    if (chu[i] > 1) root = i;
  dfs(root, 0);
  if ((!ji) || (!ou))
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  cout << n - 1 - cnt;
  return 0;
}

### Prompt

Your task is to create 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;
const int maxn = 1e5 + 10;
typedef struct a {
  int to, next;
} per;
per si[maxn * 4];
int ctt, n, chu[maxn], head[maxn], ji, ou, dep[maxn], vis[maxn], siz[maxn], cnt;
void add(int u, int v) {
  si[++ctt].next = head[u];
  si[ctt].to = v;
  head[u] = ctt;
}
int dfs(int now, int fa) {
  dep[now] = dep[fa] + 1;
  int sum = 0;
  for (int i = head[now]; i != -1; i = si[i].next) {
    int to = si[i].to;
    if (to != fa) {
      sum += dfs(to, now);
    }
  }
  if (sum > 1) cnt += sum - 1;
  if (si[head[now]].to == fa && si[head[now]].next == -1) {
    if (dep[now] & 1)
      ji = 1;
    else
      ou = 1;
    return 1;
  }
  return 0;
}
int main() {
  cin >> n;
  int u, v;
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; ++i) {
    cin >> u >> v;
    add(u, v);
    add(v, u);
    chu[u]++;
    chu[v]++;
  }
  int root;
  for (int i = 1; i <= n; ++i)
    if (chu[i] > 1) root = i;
  dfs(root, 0);
  if ((!ji) || (!ou))
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  cout << n - 1 - cnt;
  return 0;
}

```