#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long sz = 1e5 + 10;
vector<long long> g[sz];
long long now = 0, mn = 0;
bitset<10 * sz> tot;
void dfs(long long u, long long p, long long xr, long long lv) {
  bool leaf = 0;
  for (long long v : g[u]) {
    if (v == p) continue;
    if (g[v].size() != 1) {
      now++;
      tot[now] = 1;
      dfs(v, u, xr ^ 1, lv + 1);
    } else {
      if ((xr ^ 1) != 0)
        mn = max(mn, 3LL);
      else
        mn = max(mn, 1LL);
      leaf = 1;
    }
  }
  if (leaf && lv > 1) {
    now++;
    tot[now] = 1;
  }
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    scanf("%lld", &u), scanf("%lld", &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  long long leaf;
  for (long long i = 1; i <= n; i++) {
    if (g[i].size() == 1) {
      leaf = i;
      break;
    }
  }
  dfs(leaf, -1, 0, 0);
  long long mx = tot.count();
  cout << mn << " " << mx << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
#pragma GCC target("avx,avx2,fma")
using namespace std;
const long long sz = 1e5 + 10;
vector<long long> g[sz];
long long now = 0, mn = 0;
bitset<10 * sz> tot;
void dfs(long long u, long long p, long long xr, long long lv) {
  bool leaf = 0;
  for (long long v : g[u]) {
    if (v == p) continue;
    if (g[v].size() != 1) {
      now++;
      tot[now] = 1;
      dfs(v, u, xr ^ 1, lv + 1);
    } else {
      if ((xr ^ 1) != 0)
        mn = max(mn, 3LL);
      else
        mn = max(mn, 1LL);
      leaf = 1;
    }
  }
  if (leaf && lv > 1) {
    now++;
    tot[now] = 1;
  }
}
int main() {
  long long n;
  cin >> n;
  for (long long i = 1; i < n; i++) {
    long long u, v;
    scanf("%lld", &u), scanf("%lld", &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  long long leaf;
  for (long long i = 1; i <= n; i++) {
    if (g[i].size() == 1) {
      leaf = i;
      break;
    }
  }
  dfs(leaf, -1, 0, 0);
  long long mx = tot.count();
  cout << mn << " " << mx << endl;
  return 0;
}

```