#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cout << x; }
void __print(long x) { cout << x; }
void __print(long long x) { cout << x; }
void __print(unsigned x) { cout << x; }
void __print(unsigned long x) { cout << x; }
void __print(unsigned long long x) { cout << x; }
void __print(float x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
long long int fast_pow(long long int base, long long int exp,
                       long long int M = 1000000007) {
  base = base % M;
  long long int result = 1;
  while (exp > 0) {
    if (exp % 2 == 1) result = (result * base) % M;
    base = (base * base) % M;
    exp = exp / 2;
  }
  return result;
}
long long int modInverse(long long int A, long long int M = 1000000007) {
  return fast_pow(A, M - 2, M);
}
vector<vector<long long int> > adj(100000);
bool vis[100000];
long long int dpt[100000];
vector<vector<long long int> > child(100000);
void dfs(long long int src, long long int pard, long long int par) {
  dpt[src] = pard + 1;
  for (auto i : adj[src]) {
    if (i != par) {
      if (adj[i].size() == 1) {
        child[src].push_back(i);
      }
      dfs(i, dpt[src], src);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 0; i <= n - 2; i++) {
    long long int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long int root;
  for (long long int i = 0; i <= n - 1; i++) {
    if (adj[i].size() > 1) {
      root = i;
      break;
    }
  }
  bool o = false, e = false;
  dfs(root, -1, -1);
  for (long long int i = 0; i <= n - 1; i++) {
    if (adj[i].size() == 1) {
      if (dpt[i] % 2) {
        o = true;
      } else
        e = true;
    }
  }
  long long int ans = 0;
  if (o && e) {
    ans = 3;
  } else
    ans = 1;
  long long int dec = 0;
  for (long long int i = 0; i <= n - 1; i++) {
    long long int sz = child[i].size();
    if (sz > 1) dec += 1 - sz;
  }
  long long int res = n - 1 + dec;
  cout << ans << " " << res;
  return 0;
}

### Prompt

Create a solution in cpp for 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;
void __print(int x) { cout << x; }
void __print(long x) { cout << x; }
void __print(long long x) { cout << x; }
void __print(unsigned x) { cout << x; }
void __print(unsigned long x) { cout << x; }
void __print(unsigned long long x) { cout << x; }
void __print(float x) { cout << x; }
void __print(double x) { cout << x; }
void __print(long double x) { cout << x; }
void __print(char x) { cout << '\'' << x << '\''; }
void __print(const char *x) { cout << '\"' << x << '\"'; }
void __print(const string &x) { cout << '\"' << x << '\"'; }
void __print(bool x) { cout << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cout << '{';
  __print(x.first);
  cout << ',';
  __print(x.second);
  cout << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cout << '{';
  for (auto &i : x) cout << (f++ ? "," : ""), __print(i);
  cout << "}";
}
void _print() { cout << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cout << ", ";
  _print(v...);
}
long long int fast_pow(long long int base, long long int exp,
                       long long int M = 1000000007) {
  base = base % M;
  long long int result = 1;
  while (exp > 0) {
    if (exp % 2 == 1) result = (result * base) % M;
    base = (base * base) % M;
    exp = exp / 2;
  }
  return result;
}
long long int modInverse(long long int A, long long int M = 1000000007) {
  return fast_pow(A, M - 2, M);
}
vector<vector<long long int> > adj(100000);
bool vis[100000];
long long int dpt[100000];
vector<vector<long long int> > child(100000);
void dfs(long long int src, long long int pard, long long int par) {
  dpt[src] = pard + 1;
  for (auto i : adj[src]) {
    if (i != par) {
      if (adj[i].size() == 1) {
        child[src].push_back(i);
      }
      dfs(i, dpt[src], src);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  for (long long int i = 0; i <= n - 2; i++) {
    long long int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long int root;
  for (long long int i = 0; i <= n - 1; i++) {
    if (adj[i].size() > 1) {
      root = i;
      break;
    }
  }
  bool o = false, e = false;
  dfs(root, -1, -1);
  for (long long int i = 0; i <= n - 1; i++) {
    if (adj[i].size() == 1) {
      if (dpt[i] % 2) {
        o = true;
      } else
        e = true;
    }
  }
  long long int ans = 0;
  if (o && e) {
    ans = 3;
  } else
    ans = 1;
  long long int dec = 0;
  for (long long int i = 0; i <= n - 1; i++) {
    long long int sz = child[i].size();
    if (sz > 1) dec += 1 - sz;
  }
  long long int res = n - 1 + dec;
  cout << ans << " " << res;
  return 0;
}

```