#include <bits/stdc++.h>
using namespace std;
const int N = 1000005;
int T, n, a[N], cnum, tot, fl0, fl1;
long long s[N];
bool vs[N];
inline int Min(const int p, const int q) { return p < q ? p : q; }
vector<int> g[N];
inline void dfs(const int x, const int fa, const int fl) {
  if (g[x].size() == 1)
    return ++tot, cnum += (!vs[fa]), vs[fa] = 1, fl0 |= (fl == 0),
                                     fl1 |= (fl == 1), void();
  for (const int &j : g[x])
    if (j ^ fa) dfs(j, x, fl ^ 1);
}
int main() {
  scanf("%d", &n);
  register int i, u, v;
  for (i = 1; i <= n; ++i) g[i].clear(), vs[i] = 0;
  vs[0] = 0;
  tot = cnum = fl0 = fl1 = 0;
  for (i = 1; i < n; ++i)
    scanf("%d%d", &u, &v), g[u].push_back(v), g[v].push_back(u);
  for (i = 1; i <= n; ++i)
    if (g[i].size() != 1) {
      dfs(i, 0, 0);
      break;
    }
  fl0 &&fl1 ? putchar('3') : putchar('1'), printf(" %d\n", n - 1 - tot + cnum);
  return 0;
}

### Prompt

Your task is to create 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;
const int N = 1000005;
int T, n, a[N], cnum, tot, fl0, fl1;
long long s[N];
bool vs[N];
inline int Min(const int p, const int q) { return p < q ? p : q; }
vector<int> g[N];
inline void dfs(const int x, const int fa, const int fl) {
  if (g[x].size() == 1)
    return ++tot, cnum += (!vs[fa]), vs[fa] = 1, fl0 |= (fl == 0),
                                     fl1 |= (fl == 1), void();
  for (const int &j : g[x])
    if (j ^ fa) dfs(j, x, fl ^ 1);
}
int main() {
  scanf("%d", &n);
  register int i, u, v;
  for (i = 1; i <= n; ++i) g[i].clear(), vs[i] = 0;
  vs[0] = 0;
  tot = cnum = fl0 = fl1 = 0;
  for (i = 1; i < n; ++i)
    scanf("%d%d", &u, &v), g[u].push_back(v), g[v].push_back(u);
  for (i = 1; i <= n; ++i)
    if (g[i].size() != 1) {
      dfs(i, 0, 0);
      break;
    }
  fl0 &&fl1 ? putchar('3') : putchar('1'), printf(" %d\n", n - 1 - tot + cnum);
  return 0;
}

```