#include <bits/stdc++.h>
using namespace std;
inline int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
int n;
vector<int> e[200005];
map<int, int> mp;
int dep[200005], cnt0, cnt1;
void dfs(int u, int pa = 0) {
  dep[u] = dep[pa] + 1;
  for (auto v : e[u])
    if (v != pa) dfs(v, u);
}
signed main() {
  n = read();
  for (register int i = (2); i <= (n); ++i) {
    int u = read(), v = read();
    e[u].push_back(v), e[v].push_back(u);
  }
  int rt;
  for (register int i = (1); i <= (n); ++i)
    if (e[i].size() > 1) rt = i;
  dfs(rt);
  for (register int i = (1); i <= (n); ++i)
    if (e[i].size() == 1)
      mp[e[i][0]]++, cnt0 += (dep[i] & 1 ^ 1), cnt1 += (dep[i] & 1);
  int res = n - 1;
  for (auto u : mp) res -= u.second - 1;
  if (!cnt0 || !cnt1)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  cout << res;
  return 0;
}

### 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 int read() {
  char c = getchar();
  int x = 0;
  bool f = 0;
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  if (f) x = -x;
  return x;
}
int n;
vector<int> e[200005];
map<int, int> mp;
int dep[200005], cnt0, cnt1;
void dfs(int u, int pa = 0) {
  dep[u] = dep[pa] + 1;
  for (auto v : e[u])
    if (v != pa) dfs(v, u);
}
signed main() {
  n = read();
  for (register int i = (2); i <= (n); ++i) {
    int u = read(), v = read();
    e[u].push_back(v), e[v].push_back(u);
  }
  int rt;
  for (register int i = (1); i <= (n); ++i)
    if (e[i].size() > 1) rt = i;
  dfs(rt);
  for (register int i = (1); i <= (n); ++i)
    if (e[i].size() == 1)
      mp[e[i][0]]++, cnt0 += (dep[i] & 1 ^ 1), cnt1 += (dep[i] & 1);
  int res = n - 1;
  for (auto u : mp) res -= u.second - 1;
  if (!cnt0 || !cnt1)
    cout << 1 << ' ';
  else
    cout << 3 << ' ';
  cout << res;
  return 0;
}

```