#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9 + 10;
const long long int LINF = 1LL * INF * INF;
const int MAXN = 2e5 + 10;
const int MAXM = 5e3 + 10;
priority_queue<int> pq;
vector<vector<int> > graph;
queue<int> que;
int A[MAXN];
int par[MAXN];
int depth[MAXN];
int cal[MAXN];
void dfs(int here, int p) {
  int there;
  for (int i = 0; i < graph[here].size(); i++) {
    there = graph[here][i];
    if (there == p) continue;
    par[there] = here;
    depth[there] = depth[here] + 1;
    dfs(there, here);
  }
  return;
}
int main() {
  int n, m, k, a, b, x, y, q;
  int sum = 0;
  int cnt = 0;
  int mx = 0;
  int mn = INF;
  int cur = 0, idx = -1;
  int tc;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  graph = vector<vector<int> >(n + 1);
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if ((int)graph[i].size() > 1) {
      idx = i;
      break;
    }
  }
  dfs(idx, 0);
  bool judgea = true;
  bool judgeb = true;
  for (int i = 1; i <= n; i++) {
    if ((int)graph[i].size() == 1) {
      judgea &= (depth[i] & 1);
      judgeb &= (!(depth[i] & 1));
      cur = par[i];
      cal[cur]++;
    }
  }
  int res = n - 1;
  for (int i = 1; i <= n; i++) {
    if (cal[i] > 1) {
      res -= (cal[i] - 1);
    }
  }
  if (judgea || judgeb)
    cout << 1 << " " << res << "\n";
  else
    cout << 3 << " " << res << "\n";
  return 0;
}

### 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 = 1e9 + 10;
const long long int LINF = 1LL * INF * INF;
const int MAXN = 2e5 + 10;
const int MAXM = 5e3 + 10;
priority_queue<int> pq;
vector<vector<int> > graph;
queue<int> que;
int A[MAXN];
int par[MAXN];
int depth[MAXN];
int cal[MAXN];
void dfs(int here, int p) {
  int there;
  for (int i = 0; i < graph[here].size(); i++) {
    there = graph[here][i];
    if (there == p) continue;
    par[there] = here;
    depth[there] = depth[here] + 1;
    dfs(there, here);
  }
  return;
}
int main() {
  int n, m, k, a, b, x, y, q;
  int sum = 0;
  int cnt = 0;
  int mx = 0;
  int mn = INF;
  int cur = 0, idx = -1;
  int tc;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  graph = vector<vector<int> >(n + 1);
  for (int i = 1; i < n; i++) {
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  for (int i = 1; i <= n; i++) {
    if ((int)graph[i].size() > 1) {
      idx = i;
      break;
    }
  }
  dfs(idx, 0);
  bool judgea = true;
  bool judgeb = true;
  for (int i = 1; i <= n; i++) {
    if ((int)graph[i].size() == 1) {
      judgea &= (depth[i] & 1);
      judgeb &= (!(depth[i] & 1));
      cur = par[i];
      cal[cur]++;
    }
  }
  int res = n - 1;
  for (int i = 1; i <= n; i++) {
    if (cal[i] > 1) {
      res -= (cal[i] - 1);
    }
  }
  if (judgea || judgeb)
    cout << 1 << " " << res << "\n";
  else
    cout << 3 << " " << res << "\n";
  return 0;
}

```