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;
using ll = long long;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
const int MOD = 1e9 + 7;
const int dx[] = {-1, 0, 1, 0, -1, 1, 1, -1};
const int dy[] = {0, 1, 0, -1, 1, 1, -1, -1};
vector<int> adj[100000];
int id[100000];
int mn, mx;
void dfs(int now, int prv = -1) {
  int ch = 0;
  for (int nxt : adj[now])
    if (nxt != prv) {
      ++ch;
      dfs(nxt, now);
    }
  if (ch == 0 || prv == -1 && ch == 1) id[now] = 0;
}
void dfs2(int now, int prv = -1) {
  bool near = 0;
  for (int nxt : adj[now])
    if (id[nxt] == 0) near = 1;
  if (id[now] == -1) {
    if (near)
      id[now] = 1, ++mx;
    else
      id[now] = 2;
  }
  for (int nxt : adj[now])
    if (nxt != prv) {
      dfs2(nxt, now);
      if (id[now] && id[nxt]) ++mx;
    }
}
void dfs3(int now, int dist, int prv = -1) {
  if (id[now] == 0) {
    if (dist & 1) mn = 3;
  }
  for (int nxt : adj[now])
    if (nxt != prv) {
      dfs3(nxt, dist + 1, now);
    }
}
int main() {
  cin.sync_with_stdio(false);
  cin.tie(nullptr);
  memset(id, -1, sizeof(id));
  int N;
  cin >> N;
  for (int a, b, i = 1; i < N; ++i) {
    cin >> a >> b;
    adj[--a].push_back(--b);
    adj[b].push_back(a);
  }
  dfs(0);
  dfs2(0);
  mn = 1;
  for (int i = 0; i < N; ++i)
    if (id[i] == 0) {
      dfs3(i, 0);
      break;
    }
  cout << mn << ' ' << mx;
  return 0;
}
