#include <bits/stdc++.h>
using namespace std;
int p[100005];
vector<int> edges[100005];
queue<int> q;
char f[100005];
void bfs(int k) {
  q.push(k);
  f[k] = 1;
  while (!q.empty()) {
    int node = q.front();
    q.pop();
    for (auto &i : edges[node]) {
      if (edges[i].size() != 1 && f[i] == 0) {
        if (f[node] == 1)
          f[i] = 2;
        else if (f[node] == 2)
          f[i] = 1;
        q.push(i);
      }
    }
  }
}
int main() {
  int n, p1 = 0, p2 = 0, maxim, a, b;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  int noleaf = -1;
  for (int i = 1; i <= n; ++i) {
    if (edges[i].size() == 1) {
      p[edges[i][0]]++;
    } else
      noleaf = i;
  }
  if (noleaf == -1) cout << 1 << " " << 1;
  bfs(noleaf);
  for (int i = 1; i <= n; ++i) {
    if (edges[i].size() != 1) {
      if (p[i] != 0) {
        if (f[i] == 1)
          p1++;
        else if (f[i] == 2)
          p2++;
      }
    }
  }
  if (p1 == 0 || p2 == 0) {
    cout << 1 << " ";
  } else
    cout << 3 << " ";
  maxim = n - 1;
  for (int i = 1; i <= n; ++i) {
    if (edges[i].size() != 1) {
      if (p[i] >= 2) maxim -= (p[i] - 1);
    }
  }
  cout << maxim;
}

### 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;
int p[100005];
vector<int> edges[100005];
queue<int> q;
char f[100005];
void bfs(int k) {
  q.push(k);
  f[k] = 1;
  while (!q.empty()) {
    int node = q.front();
    q.pop();
    for (auto &i : edges[node]) {
      if (edges[i].size() != 1 && f[i] == 0) {
        if (f[node] == 1)
          f[i] = 2;
        else if (f[node] == 2)
          f[i] = 1;
        q.push(i);
      }
    }
  }
}
int main() {
  int n, p1 = 0, p2 = 0, maxim, a, b;
  cin >> n;
  for (int i = 1; i < n; ++i) {
    cin >> a >> b;
    edges[a].push_back(b);
    edges[b].push_back(a);
  }
  int noleaf = -1;
  for (int i = 1; i <= n; ++i) {
    if (edges[i].size() == 1) {
      p[edges[i][0]]++;
    } else
      noleaf = i;
  }
  if (noleaf == -1) cout << 1 << " " << 1;
  bfs(noleaf);
  for (int i = 1; i <= n; ++i) {
    if (edges[i].size() != 1) {
      if (p[i] != 0) {
        if (f[i] == 1)
          p1++;
        else if (f[i] == 2)
          p2++;
      }
    }
  }
  if (p1 == 0 || p2 == 0) {
    cout << 1 << " ";
  } else
    cout << 3 << " ";
  maxim = n - 1;
  for (int i = 1; i <= n; ++i) {
    if (edges[i].size() != 1) {
      if (p[i] >= 2) maxim -= (p[i] - 1);
    }
  }
  cout << maxim;
}

```