#include <bits/stdc++.h>
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const long long inf = 0x3f3f3f3f3f3f3f3f;
using namespace std;
const int N = 1e5 + 7;
int t, n, m, ans = 0, fl = 0;
vector<int> v[100005], one;
int qi = 0;
void dfs(int x, int y, int z) {
  int aa = 0;
  if (ans) return;
  if (y != -1 && v[x].size() == 1) {
    if (z % 2) ans = 1;
    return;
  }
  for (long long i = 0; i <= v[x].size() - 1; i++) {
    if (v[x][i] == y) continue;
    dfs(v[x][i], x, z + 1);
  }
  return;
}
int main() {
  scanf("%d", &n);
  int minn = 0, maxn = n - 1;
  for (long long i = 1; i <= n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    if (v[i].size() == 1 && qi == 0) {
      qi = i;
    }
    int cnt = 0;
    for (long long j = 0; j <= v[i].size() - 1; j++) {
      if (v[v[i][j]].size() == 1) cnt++;
    }
    if (cnt >= 2) fl += cnt - 1;
  }
  dfs(qi, -1, 0);
  if (ans == 0)
    minn = 1;
  else
    minn = 3;
  maxn -= fl;
  printf("%d %d\n", minn, maxn);
  return 0;
}

### Prompt

Your challenge is to write 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>
const double eps = 1e-10;
const int INF = 0x3f3f3f3f;
const long long inf = 0x3f3f3f3f3f3f3f3f;
using namespace std;
const int N = 1e5 + 7;
int t, n, m, ans = 0, fl = 0;
vector<int> v[100005], one;
int qi = 0;
void dfs(int x, int y, int z) {
  int aa = 0;
  if (ans) return;
  if (y != -1 && v[x].size() == 1) {
    if (z % 2) ans = 1;
    return;
  }
  for (long long i = 0; i <= v[x].size() - 1; i++) {
    if (v[x][i] == y) continue;
    dfs(v[x][i], x, z + 1);
  }
  return;
}
int main() {
  scanf("%d", &n);
  int minn = 0, maxn = n - 1;
  for (long long i = 1; i <= n - 1; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
    v[b].push_back(a);
  }
  for (long long i = 1; i <= n; i++) {
    if (v[i].size() == 1 && qi == 0) {
      qi = i;
    }
    int cnt = 0;
    for (long long j = 0; j <= v[i].size() - 1; j++) {
      if (v[v[i][j]].size() == 1) cnt++;
    }
    if (cnt >= 2) fl += cnt - 1;
  }
  dfs(qi, -1, 0);
  if (ans == 0)
    minn = 1;
  else
    minn = 3;
  maxn -= fl;
  printf("%d %d\n", minn, maxn);
  return 0;
}

```