#include <bits/stdc++.h>
using namespace std;
long long const INF = (1 << 30);
long long const MOD = 1000000007;
int const MAXN = 100005;
vector<int> adj[MAXN];
int dp[MAXN];
bool dpx[MAXN][2];
int f(int u, int par) {
  dp[u] = 0;
  int zero_cter = 0;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v != par) {
      int tmp = f(v, u);
      if (tmp == 0)
        zero_cter++;
      else
        dp[u] += tmp;
    }
  }
  dp[u] += adj[u].size() - 1;
  if (zero_cter > 0) dp[u] = dp[u] - zero_cter + 1;
  return dp[u];
}
bool g(int u, int par, int state) {
  dpx[u][state] = false;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    bool tmp = false;
    if (v != par) {
      tmp = g(v, u, 1 - state);
    }
    if (tmp) {
      dpx[u][state] = true;
      return dpx[u][state];
    }
  }
  if (adj[u].size() == 1 && state == 1) dpx[u][state] = false;
  if (adj[u].size() == 1 && state == 0) dpx[u][state] = true;
  return dpx[u][state];
}
int main() {
  int n, src = 0;
  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);
    if (src == 0 && adj[a].size() > 1) src = a;
  }
  adj[src].push_back(0);
  adj[0].push_back(src);
  memset(dpx, 0, sizeof dpx);
  bool tmp = g(src, 0, 1);
  bool tmp2 = g(src, 0, 0);
  if (tmp && tmp2)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  memset(dp, -1, sizeof dp);
  int tmp3 = f(src, 0);
  cout << tmp3 << "\n";
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
long long const INF = (1 << 30);
long long const MOD = 1000000007;
int const MAXN = 100005;
vector<int> adj[MAXN];
int dp[MAXN];
bool dpx[MAXN][2];
int f(int u, int par) {
  dp[u] = 0;
  int zero_cter = 0;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (v != par) {
      int tmp = f(v, u);
      if (tmp == 0)
        zero_cter++;
      else
        dp[u] += tmp;
    }
  }
  dp[u] += adj[u].size() - 1;
  if (zero_cter > 0) dp[u] = dp[u] - zero_cter + 1;
  return dp[u];
}
bool g(int u, int par, int state) {
  dpx[u][state] = false;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    bool tmp = false;
    if (v != par) {
      tmp = g(v, u, 1 - state);
    }
    if (tmp) {
      dpx[u][state] = true;
      return dpx[u][state];
    }
  }
  if (adj[u].size() == 1 && state == 1) dpx[u][state] = false;
  if (adj[u].size() == 1 && state == 0) dpx[u][state] = true;
  return dpx[u][state];
}
int main() {
  int n, src = 0;
  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);
    if (src == 0 && adj[a].size() > 1) src = a;
  }
  adj[src].push_back(0);
  adj[0].push_back(src);
  memset(dpx, 0, sizeof dpx);
  bool tmp = g(src, 0, 1);
  bool tmp2 = g(src, 0, 0);
  if (tmp && tmp2)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  memset(dp, -1, sizeof dp);
  int tmp3 = f(src, 0);
  cout << tmp3 << "\n";
  return 0;
}

```