#include <bits/stdc++.h>
using namespace std;
const int maxn = 100000 + 50;
vector<int> a[maxn];
struct node {
  int id;
  int head;
  int flag;
};
int cnt[2] = {0};
void bfs(int s) {
  queue<node> p;
  p.push({s, 0, 0});
  while (!p.empty()) {
    node t = p.front();
    p.pop();
    int i = t.id;
    if (a[i].size() == 1) cnt[t.flag]++;
    for (int j = 0; j < a[i].size(); j++) {
      if (a[i][j] != t.head) {
        p.push({a[i][j], i, t.flag ^ 1});
      }
    }
  }
}
long long max_num;
bool judge(int i) {
  if (a[i].size() == 1) return true;
  return false;
}
void bfs_max(int s) {
  queue<node> p;
  p.push({s, 0, 1});
  while (!p.empty()) {
    node t = p.front();
    p.pop();
    int i = t.id;
    int cnt = 0;
    for (int j = 0; j < a[i].size(); j++) {
      if (judge(a[i][j]))
        cnt++;
      else {
        if (a[i][j] != t.head) {
          p.push({a[i][j], i, t.flag ^ 1});
        }
      }
    }
    if (cnt > 1) max_num -= cnt - 1;
  }
}
int main() {
  long long n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    a[u].push_back(v);
    a[v].push_back(u);
  }
  max_num = n - 1;
  for (int i = 1; i <= n; i++) {
    if (a[i].size() == 1) {
      bfs(i);
      bfs_max(i);
      break;
    }
  }
  long long min_num;
  if (!(cnt[0] && cnt[1]))
    min_num = 1;
  else {
    min_num = 3;
  }
  cout << min_num << ' ' << max_num << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 maxn = 100000 + 50;
vector<int> a[maxn];
struct node {
  int id;
  int head;
  int flag;
};
int cnt[2] = {0};
void bfs(int s) {
  queue<node> p;
  p.push({s, 0, 0});
  while (!p.empty()) {
    node t = p.front();
    p.pop();
    int i = t.id;
    if (a[i].size() == 1) cnt[t.flag]++;
    for (int j = 0; j < a[i].size(); j++) {
      if (a[i][j] != t.head) {
        p.push({a[i][j], i, t.flag ^ 1});
      }
    }
  }
}
long long max_num;
bool judge(int i) {
  if (a[i].size() == 1) return true;
  return false;
}
void bfs_max(int s) {
  queue<node> p;
  p.push({s, 0, 1});
  while (!p.empty()) {
    node t = p.front();
    p.pop();
    int i = t.id;
    int cnt = 0;
    for (int j = 0; j < a[i].size(); j++) {
      if (judge(a[i][j]))
        cnt++;
      else {
        if (a[i][j] != t.head) {
          p.push({a[i][j], i, t.flag ^ 1});
        }
      }
    }
    if (cnt > 1) max_num -= cnt - 1;
  }
}
int main() {
  long long n;
  cin >> n;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    a[u].push_back(v);
    a[v].push_back(u);
  }
  max_num = n - 1;
  for (int i = 1; i <= n; i++) {
    if (a[i].size() == 1) {
      bfs(i);
      bfs_max(i);
      break;
    }
  }
  long long min_num;
  if (!(cnt[0] && cnt[1]))
    min_num = 1;
  else {
    min_num = 3;
  }
  cout << min_num << ' ' << max_num << endl;
  return 0;
}

```