#include <bits/stdc++.h>
using namespace std;
long long fastpow(long long a, long long b,
                  long long m = (long long)(1e9 + 7)) {
  long long res = 1;
  a %= m;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long n, u, v;
const long long maxN = 1e5 + 5;
vector<long long> adj[maxN];
vector<long long> paths;
void dfs(long long root, long long parent, long long level) {
  if ((long long)adj[root].size() == 1 and root != parent) {
    paths.emplace_back(level);
    return;
  }
  for (auto x : adj[root]) {
    if (x != parent) {
      dfs(x, root, level + 1);
    }
  }
}
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  vector<long long> leafs;
  for (long long i = 1; i <= n; i++) {
    if ((long long)adj[i].size() == 1) leafs.emplace_back(i);
  }
  long long m = 0, root = leafs[0];
  vector<bool> marked(n + 1);
  for (auto l : leafs) {
    if (!marked[adj[l][0]]) {
      marked[adj[l][0]] = true;
      m++;
    }
  }
  dfs(root, root, 0);
  long long o = 0;
  for (auto p : paths) {
    o += (p & 1);
  }
  if (o) {
    cout << 3 << " " << n - 1 - (long long)leafs.size() + m << endl;
  } else {
    cout << 1 << " " << n - 1 - (long long)leafs.size() + m << endl;
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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>
using namespace std;
long long fastpow(long long a, long long b,
                  long long m = (long long)(1e9 + 7)) {
  long long res = 1;
  a %= m;
  while (b > 0) {
    if (b & 1) res = (res * a) % m;
    a = (a * a) % m;
    b >>= 1;
  }
  return res;
}
long long n, u, v;
const long long maxN = 1e5 + 5;
vector<long long> adj[maxN];
vector<long long> paths;
void dfs(long long root, long long parent, long long level) {
  if ((long long)adj[root].size() == 1 and root != parent) {
    paths.emplace_back(level);
    return;
  }
  for (auto x : adj[root]) {
    if (x != parent) {
      dfs(x, root, level + 1);
    }
  }
}
int32_t main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  ;
  cin >> n;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].emplace_back(v);
    adj[v].emplace_back(u);
  }
  vector<long long> leafs;
  for (long long i = 1; i <= n; i++) {
    if ((long long)adj[i].size() == 1) leafs.emplace_back(i);
  }
  long long m = 0, root = leafs[0];
  vector<bool> marked(n + 1);
  for (auto l : leafs) {
    if (!marked[adj[l][0]]) {
      marked[adj[l][0]] = true;
      m++;
    }
  }
  dfs(root, root, 0);
  long long o = 0;
  for (auto p : paths) {
    o += (p & 1);
  }
  if (o) {
    cout << 3 << " " << n - 1 - (long long)leafs.size() + m << endl;
  } else {
    cout << 1 << " " << n - 1 - (long long)leafs.size() + m << endl;
  }
  return 0;
}

```