#include <bits/stdc++.h>
using namespace std;
const int SIZE = 123456;
vector<int> g[SIZE];
void dfs1(int pre, int cur, int depth, set<int>& st) {
  if (g[cur].size() == 1 && g[cur][0] == pre) {
    st.insert(depth % 2);
    return;
  }
  long long u = 0;
  for (auto nxt : g[cur]) {
    if (nxt == pre) {
      continue;
    }
    dfs1(cur, nxt, depth + 1, st);
  }
}
bool dfs2(int pre, int cur, int depth, int& tot) {
  if (g[cur].size() == 1 && g[cur][0] == pre) {
    return true;
  }
  int leaf = 0;
  for (auto nxt : g[cur]) {
    if (nxt == pre) {
      continue;
    }
    leaf += dfs2(cur, nxt, depth + 1, tot);
  }
  if (leaf) {
    tot -= leaf - 1;
  }
  return false;
}
int main() {
  int n;
  cin >> n;
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int root = -1;
  for (int i = 0; i < n; i++) {
    if (g[i].size() > 1) {
      root = i;
      break;
    }
  }
  assert(root != -1);
  set<int> st1;
  dfs1(-1, root, 0, st1);
  if (st1.size() == 1) {
    a = 1;
  } else {
    a = 3;
  }
  b = n - 1;
  dfs2(-1, root, 0, b);
  printf("%d %d\n", a, b);
  return 0;
}

### Prompt

Please create a solution in Cpp 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 SIZE = 123456;
vector<int> g[SIZE];
void dfs1(int pre, int cur, int depth, set<int>& st) {
  if (g[cur].size() == 1 && g[cur][0] == pre) {
    st.insert(depth % 2);
    return;
  }
  long long u = 0;
  for (auto nxt : g[cur]) {
    if (nxt == pre) {
      continue;
    }
    dfs1(cur, nxt, depth + 1, st);
  }
}
bool dfs2(int pre, int cur, int depth, int& tot) {
  if (g[cur].size() == 1 && g[cur][0] == pre) {
    return true;
  }
  int leaf = 0;
  for (auto nxt : g[cur]) {
    if (nxt == pre) {
      continue;
    }
    leaf += dfs2(cur, nxt, depth + 1, tot);
  }
  if (leaf) {
    tot -= leaf - 1;
  }
  return false;
}
int main() {
  int n;
  cin >> n;
  int a, b;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    a--;
    b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  int root = -1;
  for (int i = 0; i < n; i++) {
    if (g[i].size() > 1) {
      root = i;
      break;
    }
  }
  assert(root != -1);
  set<int> st1;
  dfs1(-1, root, 0, st1);
  if (st1.size() == 1) {
    a = 1;
  } else {
    a = 3;
  }
  b = n - 1;
  dfs2(-1, root, 0, b);
  printf("%d %d\n", a, b);
  return 0;
}

```