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;
const int mod = 1e9 + 7;
const int maxn = 1e6 + 7;
const double eps = 1e-8;
const double PI = acos(-1.0);
inline int qpow(int x, int n) {
  int ret = 1;
  for (; n; n >>= 1) {
    if (n & 1) ret = 1ll * ret * x % mod;
    x = 1ll * x * x % mod;
  }
  return ret;
}
inline int inv(int x) { return qpow(x, mod - 2); }
vector<int> G[maxn], deg;
int n, low, high;
void solve(const int cas) {
  scanf("%d", &n);
  deg.resize(n + 1, 0);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    G[u].emplace_back(v);
    G[v].emplace_back(u);
    ++deg[u];
    ++deg[v];
  }
  low = 1;
  high = n - 1;
  function<void(int, int, int)> dfs = [&](int u, int fr, int col) {
    if (deg[u] == 1 && col) low = 3;
    int has = 0;
    if (fr && deg[fr] == 1) ++has;
    for (auto v : G[u]) {
      if (v != fr) {
        dfs(v, u, col ^ 1);
        if (deg[v] == 1) ++has;
      }
    }
    --has;
    high -= max(0, has);
  };
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      dfs(i, 0, 0);
      break;
    }
  }
  printf("%d %d\n", low, high);
}
int main() {
  int T = 1, cas = 1;
  for (; T--; ++cas) solve(cas);
  return 0;
}
