#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 5;
int depth[MAXN], ind[MAXN];
struct Edge {
  int to, next;
} edges[MAXN];
int te = 0;
int par[MAXN] = {0};
inline void addE(int u, int v) {
  edges[++te].to = v, edges[te].next = par[u];
  par[u] = te;
  ind[v]++;
}
int ans = 0, n;
void dfs(int u, int fa) {
  int cnt = 0;
  for (int e = par[u]; e; e = edges[e].next) {
    int v = edges[e].to;
    if (v != fa) {
      depth[v] = depth[u] + 1;
      dfs(v, u);
    }
    if (ind[v] == 1) cnt++;
  }
  if (cnt > 0) ans -= (cnt - 1);
}
int st;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    addE(u, v), addE(v, u);
  }
  for (int i = 1; i <= n; ++i) {
    if (ind[i] >= 2) {
      st = i;
      break;
    }
  }
  if (!st) {
    cout << 1 << endl;
    return 0;
  }
  dfs(st, -1);
  bool mark0 = 0, mark1 = 0;
  for (int i = 1; i <= n; ++i) {
    if (ind[i] == 1) {
      if (depth[i] & 1)
        mark1 = 1;
      else
        mark0 = 1;
    }
  }
  if (mark1 && mark0)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << ans + n - 1 << endl;
  return 0;
}

### Prompt

Create a solution in cpp for 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 MAXN = 2e5 + 5;
int depth[MAXN], ind[MAXN];
struct Edge {
  int to, next;
} edges[MAXN];
int te = 0;
int par[MAXN] = {0};
inline void addE(int u, int v) {
  edges[++te].to = v, edges[te].next = par[u];
  par[u] = te;
  ind[v]++;
}
int ans = 0, n;
void dfs(int u, int fa) {
  int cnt = 0;
  for (int e = par[u]; e; e = edges[e].next) {
    int v = edges[e].to;
    if (v != fa) {
      depth[v] = depth[u] + 1;
      dfs(v, u);
    }
    if (ind[v] == 1) cnt++;
  }
  if (cnt > 0) ans -= (cnt - 1);
}
int st;
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    addE(u, v), addE(v, u);
  }
  for (int i = 1; i <= n; ++i) {
    if (ind[i] >= 2) {
      st = i;
      break;
    }
  }
  if (!st) {
    cout << 1 << endl;
    return 0;
  }
  dfs(st, -1);
  bool mark0 = 0, mark1 = 0;
  for (int i = 1; i <= n; ++i) {
    if (ind[i] == 1) {
      if (depth[i] & 1)
        mark1 = 1;
      else
        mark0 = 1;
    }
  }
  if (mark1 && mark0)
    cout << 3 << " ";
  else
    cout << 1 << " ";
  cout << ans + n - 1 << endl;
  return 0;
}

```