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 = (int)1.01e9;
const long long infll = (long long)1.01e18;
const long double eps = 1e-9;
const long double pi = acos((long double)-1);
mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count());
int rnd(int x) { return mrand() % x; }
void precalc() {}
const int maxn = (int)1e5 + 5;
int n;
vector<int> g[maxn];
bool read() {
  if (scanf("%d", &n) < 1) {
    return false;
  }
  for (int i = 0; i < n; i++) {
    g[i].clear();
  }
  for (int i = 0; i < n - 1; i++) {
    int v, u;
    scanf("%d%d", &v, &u);
    v--;
    u--;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  return true;
}
int col[maxn];
void dfs(int v, int p) {
  for (int i = 0; i < ((int)(g[v]).size()); i++) {
    int u = g[v][i];
    if (u == p) {
      continue;
    }
    col[u] = (col[v] ^ 1);
    dfs(u, v);
  }
}
int used[maxn];
void solve() {
  col[0] = 0;
  dfs(0, -1);
  int cols = 0;
  for (int v = 0; v < n; v++) {
    if (((int)(g[v]).size()) != 1) {
      continue;
    }
    cols |= (1 << col[v]);
  }
  int mn = (cols == 3 ? 3 : 1);
  int mx = n - 1;
  for (int v = 0; v < n; v++) {
    used[v] = false;
  }
  for (int v = 0; v < n; v++) {
    if (((int)(g[v]).size()) != 1) {
      continue;
    }
    int u = g[v][0];
    if (used[u]) {
      mx--;
    }
    used[u] = true;
  }
  printf("%d %d\n", mn, mx);
}
int main() {
  precalc();
  while (read()) {
    solve();
  }
  return 0;
}
