#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
const int mod = 1e9 + 7;
const ll inf = ll(0x3f3f3f3f3f3f3f3f);
const double PI = acos(-1), eps = 1e-9;
const int mxn = 1e5 + 5, mxa = 1e5, mxk = 2e3;
int n, dis = -1, c1 = 0, c2 = 0;
vi adj[mxn];
bool l[mxn], ok = 1;
void dfs(int v, int d = 0, int p = -1) {
  if ((int)(adj[v]).size() == 1) {
    l[v] = 1;
    if (dis == -1 && v != -1)
      dis = d;
    else if (dis != d && v != -1)
      ok = 0;
  }
  for (const int x : adj[v]) {
    if (x != p) dfs(x, d ^ 1, v);
  }
}
void dfs2(int v, int p = -1) {
  if (l[v]) ++c1;
  bool connected = 0;
  for (const int x : adj[v]) {
    connected |= l[x];
    if (x != p) dfs2(x, v);
  }
  if (!l[v] && connected) ++c2;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1);
  dfs2(1);
  cout << (ok ? 1 : 3) << ' ';
  cout << ((n - 1) - c1 + c2) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  while (t-- > 0) solve();
}

### Prompt

Generate 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;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
const int mod = 1e9 + 7;
const ll inf = ll(0x3f3f3f3f3f3f3f3f);
const double PI = acos(-1), eps = 1e-9;
const int mxn = 1e5 + 5, mxa = 1e5, mxk = 2e3;
int n, dis = -1, c1 = 0, c2 = 0;
vi adj[mxn];
bool l[mxn], ok = 1;
void dfs(int v, int d = 0, int p = -1) {
  if ((int)(adj[v]).size() == 1) {
    l[v] = 1;
    if (dis == -1 && v != -1)
      dis = d;
    else if (dis != d && v != -1)
      ok = 0;
  }
  for (const int x : adj[v]) {
    if (x != p) dfs(x, d ^ 1, v);
  }
}
void dfs2(int v, int p = -1) {
  if (l[v]) ++c1;
  bool connected = 0;
  for (const int x : adj[v]) {
    connected |= l[x];
    if (x != p) dfs2(x, v);
  }
  if (!l[v] && connected) ++c2;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1);
  dfs2(1);
  cout << (ok ? 1 : 3) << ' ';
  cout << ((n - 1) - c1 + c2) << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  while (t-- > 0) solve();
}

```