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>
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const long long LNF = 0x3f3f3f3f3f3f3f3f;
const int MOD = 1000000007;
const double EPS = 1e-8;
const long double EUL = 2.71828182845904523536;
const long double PII = 3.14159265358979323846;
vector<vector<int>> graph;
vector<int> level;
void dfs(int u, int p) {
  for (auto v : graph[u]) {
    if (v == p) continue;
    level[v] = level[u] + 1;
    dfs(v, u);
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  while (cin >> n) {
    graph.assign(n + 1, vector<int>());
    level.assign(n + 1, 0);
    for (int i = 1; i < n; i++) {
      int a, b;
      cin >> a >> b;
      graph[a].push_back(b);
      graph[b].push_back(a);
    }
    int root;
    for (int i = 1; i <= n; i++)
      if (graph[i].size() == 1) {
        root = i;
        break;
      }
    dfs(root, 0);
    int mn, mx;
    bool bijor = false;
    for (int i = 1; i <= n; i++)
      if (graph[i].size() == 1 && level[i] & 1) {
        bijor = true;
        break;
      }
    if (bijor)
      mn = 3;
    else
      mn = 1;
    mx = n - 1;
    vector<int> visited(n + 1, 0);
    for (int i = 1; i <= n; i++)
      if (graph[i].size() == 1) {
        if (!visited[graph[i][0]])
          visited[graph[i][0]] = true;
        else
          mx--;
      }
    cout << mn << " " << mx << endl;
  }
  cerr << "Time : " << 1000 * (long double)clock() / (long double)CLOCKS_PER_SEC
       << "ms\n";
  return 0;
}
