#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const int N = 1e5 + 1;
vector<int> adj[N];
vector<int> color(N);
vector<int> vis(N);
vector<int> parent[N];
void dfs(int v, int p) {
  vis[v] = 1;
  for (auto child : adj[v]) {
    if (!vis[child]) {
      color[child] = color[v] ^ 1;
      dfs(child, v);
    }
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v, --u, --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0, -1);
  set<int> leaves;
  for (int i = 0; i < n; i++) {
    if (adj[i].size() == 1) {
      leaves.insert(i);
      int p = adj[i][0];
      parent[p].push_back(i);
    }
  }
  int reds = 0, blacks = 0;
  int lsize = leaves.size();
  for (auto leaf : leaves) {
    if (color[leaf] == 0)
      reds++;
    else
      blacks++;
  }
  int mini, maxi;
  maxi = n - 1;
  for (int i = 0; i < n; i++) {
    if (parent[i].size() == 0)
      continue;
    else {
      maxi -= (parent[i].size() - 1);
    }
  }
  if (blacks == 0 || reds == 0)
    mini = 1;
  else
    mini = 3;
  cout << mini << " " << maxi << endl;
}
int main() { solve(); }

### Prompt

Please create a solution in CPP to 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>
using namespace std;
const long long mod = 1000000007;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
const int N = 1e5 + 1;
vector<int> adj[N];
vector<int> color(N);
vector<int> vis(N);
vector<int> parent[N];
void dfs(int v, int p) {
  vis[v] = 1;
  for (auto child : adj[v]) {
    if (!vis[child]) {
      color[child] = color[v] ^ 1;
      dfs(child, v);
    }
  }
}
void solve() {
  int n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v, --u, --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0, -1);
  set<int> leaves;
  for (int i = 0; i < n; i++) {
    if (adj[i].size() == 1) {
      leaves.insert(i);
      int p = adj[i][0];
      parent[p].push_back(i);
    }
  }
  int reds = 0, blacks = 0;
  int lsize = leaves.size();
  for (auto leaf : leaves) {
    if (color[leaf] == 0)
      reds++;
    else
      blacks++;
  }
  int mini, maxi;
  maxi = n - 1;
  for (int i = 0; i < n; i++) {
    if (parent[i].size() == 0)
      continue;
    else {
      maxi -= (parent[i].size() - 1);
    }
  }
  if (blacks == 0 || reds == 0)
    mini = 1;
  else
    mini = 3;
  cout << mini << " " << maxi << endl;
}
int main() { solve(); }

```