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;
int sub[100005], dis[100005], edges[100005], ans;
bool vis[100005];
void dfs(vector<int> v[], int index, int parent) {
  int pos;
  sub[index] = 1;
  for (unsigned int i = 0; i < v[index].size(); i++) {
    pos = v[index][i];
    if (pos != parent) {
      dfs(v, pos, index);
      sub[index] = sub[index] + sub[pos];
    }
  }
}
void dfs1(vector<int> v[], int index, int parent) {
  int pos, cnt = 0;
  for (unsigned int i = 0; i < v[index].size(); i++) {
    pos = v[index][i];
    if (pos != parent && sub[pos] == 1)
      cnt++;
    else if (pos != parent)
      dfs1(v, pos, index);
  }
  if (cnt > 0) ans = ans - (cnt - 1);
}
void bfs(vector<int> v[], int root) {
  queue<int> q;
  q.push(root);
  dis[root] = 0;
  while (!q.empty()) {
    int s = q.front();
    vis[s] = 1;
    q.pop();
    for (unsigned int i = 0; i < v[s].size(); i++) {
      int pos = v[s][i];
      if (!vis[pos]) {
        dis[pos] = dis[s] + 1;
        q.push(pos);
      }
    }
  }
}
int main() {
  int n, a, b;
  cin >> n;
  ans = n - 1;
  vector<int> v[n];
  for (int i = 0; i < n - 1; i++) {
    cin >> a >> b;
    edges[a - 1]++, edges[b - 1]++;
    v[a - 1].push_back(b - 1);
    v[b - 1].push_back(a - 1);
  }
  int ed = -1, root;
  for (int i = 0; i < n; i++) {
    if (edges[i] > ed) {
      ed = edges[i];
      root = i;
    }
  }
  dfs(v, root, -1);
  dfs1(v, root, -1);
  bfs(v, root);
  bool first = false, odd = false, flag = false;
  for (int i = 0; i < n; i++) {
    if (flag) break;
    if (sub[i] == 1) {
      if (!first) {
        first = true;
        if ((dis[i] % 2) == 1) odd = true;
      } else {
        if ((dis[i] % 2) == 1) {
          if (!odd) flag = true;
        } else {
          if (odd) flag = true;
        }
      }
    }
  }
  if (flag)
    cout << 3 << " " << ans << endl;
  else
    cout << 1 << " " << ans << endl;
}
