#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using PII = pair<int, int>;
const int mod = 998244353;
const int inf = 1 << 30;
const int maxn = 200000 + 5;
int n, mx = 0, mn = 1, dep[maxn];
vector<int> edge[maxn];
bool isLeaf(int u) { return (int)edge[u].size() == 1; }
void dfs(int u, int f) {
  bool flag = false;
  for (int v : edge[u]) {
    if (v == f) continue;
    dep[v] = dep[u] + 1;
    if (isLeaf(v)) {
      flag = true;
    } else {
      mx++;
      dfs(v, u);
    }
  }
  if (flag && dep[u] > 1) {
    mx++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2, u, v; i <= n; i++) {
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  int rt = -1;
  for (int i = 1; i <= n; i++) {
    if (isLeaf(i)) {
      rt = i;
      break;
    }
  }
  dfs(rt, -1);
  for (int i = 1; i <= n; i++) {
    if (isLeaf(i) && dep[i] % 2) {
      mn = 3;
      break;
    }
  }
  printf("%d %d\n", mn, mx);
  return 0;
}

### Prompt

Please create a solution in CPP 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;
using ll = long long;
using PII = pair<int, int>;
const int mod = 998244353;
const int inf = 1 << 30;
const int maxn = 200000 + 5;
int n, mx = 0, mn = 1, dep[maxn];
vector<int> edge[maxn];
bool isLeaf(int u) { return (int)edge[u].size() == 1; }
void dfs(int u, int f) {
  bool flag = false;
  for (int v : edge[u]) {
    if (v == f) continue;
    dep[v] = dep[u] + 1;
    if (isLeaf(v)) {
      flag = true;
    } else {
      mx++;
      dfs(v, u);
    }
  }
  if (flag && dep[u] > 1) {
    mx++;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 2, u, v; i <= n; i++) {
    scanf("%d%d", &u, &v);
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  int rt = -1;
  for (int i = 1; i <= n; i++) {
    if (isLeaf(i)) {
      rt = i;
      break;
    }
  }
  dfs(rt, -1);
  for (int i = 1; i <= n; i++) {
    if (isLeaf(i) && dep[i] % 2) {
      mn = 3;
      break;
    }
  }
  printf("%d %d\n", mn, mx);
  return 0;
}

```