#include <bits/stdc++.h>
using namespace std;
const int Inf = 2e9;
const int MaxN = 1e5 + 5;
int n;
int Max;
int Min;
vector<int> G[MaxN];
void DFS_Max(int x, int p = -1) {
  for (auto v : G[x]) {
    if (v == p) continue;
    DFS_Max(v, x);
  }
  int cnt = 0;
  for (auto v : G[x]) {
    if (v == p) continue;
    if (G[v].size() == 1) cnt++;
  }
  if (cnt > 1) Max -= cnt - 1;
}
vector<int> O;
vector<int> E;
void DFS_Min(int x, int p = -1) {
  if (G[x].size() == 1) {
    E[x] = 1;
    return;
  }
  for (auto v : G[x]) {
    if (v == p) continue;
    DFS_Min(v, x);
  }
  int o = 0;
  int e = 0;
  for (auto v : G[x]) {
    if (v == p) continue;
    if (E[v]) o++;
    if (O[v]) e++;
    E[x] += O[v];
    O[x] += E[v];
  }
  if (e == 0 || o == 0) return;
  Min = 3;
  return;
  for (auto v : G[x]) {
    if (v == p) continue;
    if (O[v]) {
      Min += o - (E[v] > 0);
    }
  }
}
int Go() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int root = 1;
  Max = n - 1;
  Min = 1;
  for (int i = 1; i <= n; ++i)
    if (G[i].size() > 1) root = i;
  DFS_Max(root);
  O.resize(n + 1, 0);
  E.resize(n + 1, 0);
  DFS_Min(root);
  cout << Min << " " << Max << endl;
  return 0;
}
int main() {
  int t;
  if (0)
    scanf("%d", &t);
  else
    t = 1;
  while (t--) Go();
}

### 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 Inf = 2e9;
const int MaxN = 1e5 + 5;
int n;
int Max;
int Min;
vector<int> G[MaxN];
void DFS_Max(int x, int p = -1) {
  for (auto v : G[x]) {
    if (v == p) continue;
    DFS_Max(v, x);
  }
  int cnt = 0;
  for (auto v : G[x]) {
    if (v == p) continue;
    if (G[v].size() == 1) cnt++;
  }
  if (cnt > 1) Max -= cnt - 1;
}
vector<int> O;
vector<int> E;
void DFS_Min(int x, int p = -1) {
  if (G[x].size() == 1) {
    E[x] = 1;
    return;
  }
  for (auto v : G[x]) {
    if (v == p) continue;
    DFS_Min(v, x);
  }
  int o = 0;
  int e = 0;
  for (auto v : G[x]) {
    if (v == p) continue;
    if (E[v]) o++;
    if (O[v]) e++;
    E[x] += O[v];
    O[x] += E[v];
  }
  if (e == 0 || o == 0) return;
  Min = 3;
  return;
  for (auto v : G[x]) {
    if (v == p) continue;
    if (O[v]) {
      Min += o - (E[v] > 0);
    }
  }
}
int Go() {
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int root = 1;
  Max = n - 1;
  Min = 1;
  for (int i = 1; i <= n; ++i)
    if (G[i].size() > 1) root = i;
  DFS_Max(root);
  O.resize(n + 1, 0);
  E.resize(n + 1, 0);
  DFS_Min(root);
  cout << Min << " " << Max << endl;
  return 0;
}
int main() {
  int t;
  if (0)
    scanf("%d", &t);
  else
    t = 1;
  while (t--) Go();
}

```