#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
inline void wri(long long x) {
  write(x);
  putchar(' ');
}
const int maxn = 1e5 + 233;
bool vis[3];
vector<int> edge[maxn];
int deg[maxn], n;
int ans0, f[maxn];
int dfs(int u, int fa, int d) {
  f[u] = 0;
  if (deg[u] == 1) {
    vis[d] = true;
    return 1;
  }
  int flag = 0;
  for (auto v : edge[u]) {
    if (v == fa) continue;
    if (dfs(v, u, d ^ 1))
      flag = 1;
    else
      f[u] += f[v] + 1;
  }
  f[u] += flag;
  return 0;
}
int main() {
  n = read();
  for (register int i = (1); i <= int(n); ++i) deg[i] = 0;
  for (register int i = (1); i < int(n); ++i) {
    int x = read(), y = read();
    edge[x].push_back(y);
    edge[y].push_back(x);
    deg[x]++, deg[y]++;
  }
  for (register int i = (1); i <= int(n); ++i)
    if (deg[i] >= 2) {
      vis[0] = vis[1] = false;
      ans0 = 0;
      dfs(i, 0, 0);
      if (vis[0] && vis[1])
        ans0 = 3;
      else
        ans0 = 1;
      wri(ans0), writeln(f[i]);
      return 0;
    }
  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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
  return x * f;
}
inline void write(long long x) {
  if (x < 0) x = -x, putchar('-');
  if (x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}
inline void writeln(long long x) {
  write(x);
  putchar('\n');
}
inline void wri(long long x) {
  write(x);
  putchar(' ');
}
const int maxn = 1e5 + 233;
bool vis[3];
vector<int> edge[maxn];
int deg[maxn], n;
int ans0, f[maxn];
int dfs(int u, int fa, int d) {
  f[u] = 0;
  if (deg[u] == 1) {
    vis[d] = true;
    return 1;
  }
  int flag = 0;
  for (auto v : edge[u]) {
    if (v == fa) continue;
    if (dfs(v, u, d ^ 1))
      flag = 1;
    else
      f[u] += f[v] + 1;
  }
  f[u] += flag;
  return 0;
}
int main() {
  n = read();
  for (register int i = (1); i <= int(n); ++i) deg[i] = 0;
  for (register int i = (1); i < int(n); ++i) {
    int x = read(), y = read();
    edge[x].push_back(y);
    edge[y].push_back(x);
    deg[x]++, deg[y]++;
  }
  for (register int i = (1); i <= int(n); ++i)
    if (deg[i] >= 2) {
      vis[0] = vis[1] = false;
      ans0 = 0;
      dfs(i, 0, 0);
      if (vis[0] && vis[1])
        ans0 = 3;
      else
        ans0 = 1;
      wri(ans0), writeln(f[i]);
      return 0;
    }
  return 0;
}

```