#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const long long M = 1e9 + 7;
const long long inf = 1e18 + 5;
vector<int> g[N];
int level[N];
void dfs(int node, int par) {
  for (int it : g[node]) {
    if (it != par) {
      level[it] = level[node] + 1;
      dfs(it, node);
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x;
    scanf("%d", &x);
    int y;
    scanf("%d", &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int root = -1;
  for (int i = 1; i <= n; i++) {
    if ((int)g[i].size() == 1) {
      root = i;
      break;
    }
  }
  level[root] = 0;
  dfs(root, -1);
  int flag = 0;
  for (int i = 1; i <= n; i++) {
    if ((int)g[i].size() == 1 and level[i] & 1) {
      flag = 1;
      break;
    }
  }
  int ansmin, ansmax;
  if (flag)
    ansmin = 3;
  else
    ansmin = 1;
  ansmax = n - 1;
  int vis[N] = {0};
  for (int i = 1; i <= n; i++) {
    if ((int)g[i].size() == 1) {
      if (!vis[g[i][0]])
        vis[g[i][0]] = 1;
      else
        ansmax--;
    }
  }
  printf("%d %d\n", ansmin, ansmax);
}

### 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>
using namespace std;
const int N = 1e5 + 5;
const long long M = 1e9 + 7;
const long long inf = 1e18 + 5;
vector<int> g[N];
int level[N];
void dfs(int node, int par) {
  for (int it : g[node]) {
    if (it != par) {
      level[it] = level[node] + 1;
      dfs(it, node);
    }
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x;
    scanf("%d", &x);
    int y;
    scanf("%d", &y);
    g[x].push_back(y);
    g[y].push_back(x);
  }
  int root = -1;
  for (int i = 1; i <= n; i++) {
    if ((int)g[i].size() == 1) {
      root = i;
      break;
    }
  }
  level[root] = 0;
  dfs(root, -1);
  int flag = 0;
  for (int i = 1; i <= n; i++) {
    if ((int)g[i].size() == 1 and level[i] & 1) {
      flag = 1;
      break;
    }
  }
  int ansmin, ansmax;
  if (flag)
    ansmin = 3;
  else
    ansmin = 1;
  ansmax = n - 1;
  int vis[N] = {0};
  for (int i = 1; i <= n; i++) {
    if ((int)g[i].size() == 1) {
      if (!vis[g[i][0]])
        vis[g[i][0]] = 1;
      else
        ansmax--;
    }
  }
  printf("%d %d\n", ansmin, ansmax);
}

```