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;
const int N = 1e6 + 10;
const long long mod = 1e9 + 7;
template <typename T>
inline T gcd(T a, T b) {
  return !b ? a : gcd(b, a % b);
}
template <typename T>
inline T q_pow(T a, T x) {
  T ans = 1, tmp = a;
  while (x) {
    if (x & 1) (ans *= tmp) %= mod;
    (tmp *= tmp) %= mod;
    x >>= 1;
  }
  return ans;
}
template <typename T>
inline void re(T &N) {
  int f = 1;
  char c;
  while ((c = getchar()) < '0' || c > '9')
    if (c == '-') f = -1;
  N = c - '0';
  while ((c = getchar()) >= '0' && c <= '9') N = N * 10 + c - '0';
  N *= f;
}
int m, n, t = 1, st, en, ans;
struct edge {
  int to, nxt;
  edge(int to = 0, int nxt = 0) : to(to), nxt(nxt) {}
} e[N << 1];
int head[N];
int s[N], fa[N], son[N], d[N], id[N], top[N], cnt, q[N], r[N];
inline void add(int x, int y) {
  e[++t] = edge(y, head[x]);
  head[x] = t;
  e[++t] = edge(x, head[y]);
  head[y] = t;
}
void dfs1(int x, int p) {
  s[x] = 1;
  fa[x] = p;
  int ma = -1;
  d[x] = d[fa[x]] + 1;
  int sum = 0;
  for (int i = head[x]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to == p) continue;
    dfs1(to, x);
    s[x] += s[to];
    if (s[to] == 1) sum++;
    if (s[to] > ma) ma = s[to], son[x] = to;
  }
  if (ma == -1) q[++q[0]] = x;
  if (sum > 1) ans -= sum - 1;
}
void dfs2(int x, int topf) {
  top[x] = topf;
  if (son[x]) dfs2(son[x], topf);
  for (int i = head[x]; i; i = e[i].nxt) {
    int to = e[i].to;
    if (to == fa[x] || to == son[x]) continue;
    dfs2(to, to);
  }
}
inline long long range_q(int x, int y) {
  while (top[x] != top[y]) {
    if (d[top[x]] < d[top[y]]) swap(x, y);
    x = fa[top[x]];
  }
  if (d[x] > d[y]) swap(x, y);
  return x;
}
int main() {
  re(n);
  for (int i = 1; i < n; i++) {
    int x, y;
    re(x);
    re(y);
    add(x, y);
    r[x]++;
    r[y]++;
  }
  int rt = 0;
  for (int i = 1; i <= n; i++)
    if (r[i] > 1) {
      rt = i;
      break;
    }
  ans = n - 1;
  dfs1(rt, 0);
  dfs2(rt, rt);
  int flag = 0;
  for (int i = 2; i <= q[0]; i++) {
    int lca = range_q(q[i - 1], q[i]);
    int d1 = d[q[i - 1]] - d[lca];
    int d2 = d[q[i]] - d[lca];
    if ((d1 + d2) & 1) flag = 1;
  }
  printf("%d %d\n", flag ? 3 : 1, ans);
  return 0;
}
