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>
int dp[100001], n, dep[100001], deg[100001];
bool b[2];
struct edge {
  int v, next;
  edge() {}
  edge(int _v, int _next) {
    v = _v;
    next = _next;
  }
} e[200001];
int head[100001], sz;
void init() {
  memset(head, -1, sizeof(head));
  sz = 0;
}
void insert(int u, int v) {
  ++deg[u];
  e[++sz] = edge(v, head[u]);
  head[u] = sz;
}
void dfs(int u, int f) {
  bool lf = false;
  dep[u] = dep[f] + 1;
  for (int i = head[u]; ~i; i = e[i].next) {
    if (e[i].v == f) continue;
    dfs(e[i].v, u);
    if (deg[e[i].v] == 1)
      lf = true;
    else
      dp[u] += dp[e[i].v], ++dp[u];
  }
  dp[u] += lf;
}
int main() {
  scanf("%d", &n);
  init();
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    insert(u, v);
    insert(v, u);
  }
  int s = 0;
  for (int i = 1; i <= n; ++i)
    if (deg[i] > 1) s = i;
  dfs(s, 0);
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 1) {
      b[dep[i] & 1] = true;
    }
  }
  printf("%d %d\n", b[0] && b[1] ? 3 : 1, dp[s]);
  return 0;
}
