#include <bits/stdc++.h>
using namespace std;
void err(istream_iterator<string> it) {
  cout << '\n';
  ;
}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ", ";
  err(++it, args...);
}
template <typename T>
inline void Int(T &n) {
  n = 0;
  int f = 1;
  register int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T, typename TT>
inline void Int(T &n, TT &m) {
  Int(n);
  Int(m);
}
template <typename T, typename TT, typename TTT>
inline void Int(T &n, TT &m, TTT &l) {
  Int(n, m);
  Int(l);
}
double ind() {
  double x;
  scanf("%lf", &x);
  return x;
}
int set_1(int n, int pos) { return n = (n | (1 << pos)); }
int reset_0(int n, int pos) { return n = n & ~(1 << pos); }
bool check_bit(int n, int pos) { return n = n & (1 << pos); }
int dx[8] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[8] = {0, 1, 0, -1, -1, 1, -1, 1};
const int N = (int)1e6 + 5;
const int M = (int)1e9 + 7;
const double pi = 2 * acos(0.0);
const double eps = 1e-9;
std::vector<int> adj[N];
int deg[N];
int ans[N], dis[N];
int leaf = 0, have_leaf = 0;
void dfs(int u, int p, int d) {
  if (adj[u].size() == 1) {
    leaf++;
  }
  bool f = false;
  for (auto v : adj[u]) {
    if (v != p) {
      dis[v] = dis[u] + 1;
      dfs(v, u, d + 1);
      if (deg[v] == 1) {
        f = true;
      }
    }
  }
  if (f) {
    have_leaf++;
  }
}
int main() {
  int n;
  Int(n);
  int root;
  for (int i = 1; i < n; ++i) {
    int u, v;
    Int(u, v);
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++, deg[v]++;
    if (deg[u] > 1) root = u;
    if (deg[v] > 1) root = v;
  }
  dfs(root, root, 0);
  int even = 0, odd = 0;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      even += (dis[i] % 2 == 0);
      odd += (dis[i] % 2 == 1);
    }
  }
  if (even != 0 and odd != 0) {
    printf("3 ");
  } else {
    printf("1 ");
  }
  printf("%d\n", n - 1 - leaf + have_leaf);
  return 0;
}

### Prompt

Please formulate a cpp solution 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;
void err(istream_iterator<string> it) {
  cout << '\n';
  ;
}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << ", ";
  err(++it, args...);
}
template <typename T>
inline void Int(T &n) {
  n = 0;
  int f = 1;
  register int ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) n = (n << 3) + (n << 1) + ch - '0';
  n = n * f;
}
template <typename T, typename TT>
inline void Int(T &n, TT &m) {
  Int(n);
  Int(m);
}
template <typename T, typename TT, typename TTT>
inline void Int(T &n, TT &m, TTT &l) {
  Int(n, m);
  Int(l);
}
double ind() {
  double x;
  scanf("%lf", &x);
  return x;
}
int set_1(int n, int pos) { return n = (n | (1 << pos)); }
int reset_0(int n, int pos) { return n = n & ~(1 << pos); }
bool check_bit(int n, int pos) { return n = n & (1 << pos); }
int dx[8] = {1, 0, -1, 0, -1, -1, 1, 1};
int dy[8] = {0, 1, 0, -1, -1, 1, -1, 1};
const int N = (int)1e6 + 5;
const int M = (int)1e9 + 7;
const double pi = 2 * acos(0.0);
const double eps = 1e-9;
std::vector<int> adj[N];
int deg[N];
int ans[N], dis[N];
int leaf = 0, have_leaf = 0;
void dfs(int u, int p, int d) {
  if (adj[u].size() == 1) {
    leaf++;
  }
  bool f = false;
  for (auto v : adj[u]) {
    if (v != p) {
      dis[v] = dis[u] + 1;
      dfs(v, u, d + 1);
      if (deg[v] == 1) {
        f = true;
      }
    }
  }
  if (f) {
    have_leaf++;
  }
}
int main() {
  int n;
  Int(n);
  int root;
  for (int i = 1; i < n; ++i) {
    int u, v;
    Int(u, v);
    adj[u].push_back(v);
    adj[v].push_back(u);
    deg[u]++, deg[v]++;
    if (deg[u] > 1) root = u;
    if (deg[v] > 1) root = v;
  }
  dfs(root, root, 0);
  int even = 0, odd = 0;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      even += (dis[i] % 2 == 0);
      odd += (dis[i] % 2 == 1);
    }
  }
  if (even != 0 and odd != 0) {
    printf("3 ");
  } else {
    printf("1 ");
  }
  printf("%d\n", n - 1 - leaf + have_leaf);
  return 0;
}

```