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>
#include <bits/stdc++.h>
using namespace std;
inline char gc() {
  static const int L = 233333;
  static char sxd[L], *sss = sxd, *ttt = sxd;
  if (sss == ttt) {
    ttt = (sss = sxd) + fread(sxd, 1, L, stdin);
    if (sss == ttt) {
      return EOF;
    }
  }
  return *sss++;
}
inline char readalpha() {
  char c = gc();
  for (; !isalpha(c); c = gc())
    ;
  return c;
}
inline char readchar() {
  char c = gc();
  for (; c == ' '; c = gc())
    ;
  return c;
}
template <class T>
inline bool read(T& x) {
  bool flg = false;
  char c = gc();
  x = 0;
  for (; !isdigit(c); c = gc()) {
    if (c == '-') {
      flg = true;
    } else if (c == EOF) {
      return false;
    }
  }
  for (; isdigit(c); c = gc()) {
    x = (x * 10) + (c ^ 48);
  }
  if (flg) {
    x = -x;
  }
  return true;
}
template <class T>
inline void write(T x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  if (x < 10) {
    putchar(x | 48);
    return;
  }
  write(x / 10);
  putchar((x % 10) | 48);
}
template <class T>
inline void writesp(T x) {
  write(x);
  putchar(' ');
}
template <class T>
inline void writeln(T x) {
  write(x);
  puts("");
}
namespace dfcmd {
const int maxn = 100005;
const int inf = 0x3f3f3f3f;
int n;
struct Edge {
  int to, nxt;
} e[maxn << 1];
int du[maxn];
int first[maxn];
inline void add_edge(int u, int v) {
  static int cnt = 0;
  du[u]++, du[v]++;
  e[++cnt].nxt = first[u];
  first[u] = cnt;
  e[cnt].to = v;
  e[++cnt].nxt = first[v];
  first[v] = cnt;
  e[cnt].to = u;
}
int ansalb = 0;
int flg = 1;
inline int dfs(int now, int fa) {
  int alb = 0;
  int nowans = -1;
  for (int i = first[now]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to != fa) {
      int albb = dfs(to, now);
      if (du[to] == 1) {
        alb++;
      }
      albb ^= 1;
      if (nowans == -1) {
        nowans = albb;
      } else if (nowans != albb) {
        flg = 3;
      }
    }
  }
  if (alb) {
    alb--;
  }
  ansalb += alb;
  if (nowans == -1) {
    nowans = 0;
  }
  return nowans;
}
int main() {
  read(n);
  for (int i = 1, u, v; i < n; ++i) {
    read(u), read(v);
    add_edge(u, v);
  }
  int rt = 1;
  for (int i = 1; i <= n; ++i) {
    if (du[i] >= 2) {
      rt = i;
      break;
    }
  }
  dfs(rt, 0);
  writesp(flg);
  writeln(n - ansalb - 1);
  return 0;
}
}  // namespace dfcmd
int main() { return dfcmd::main(); }
