#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long double pi = 3.1415926535897932384626433;
const long long inf = 1e18;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = ((res % mod) * (x % mod)) % mod;
    y = y >> 1;
    x = ((x % mod) * (x % mod)) % mod;
  }
  return res;
}
long long max(long long a, long long b) {
  long long ans = a > b ? a : b;
  return ans;
}
vector<vector<long long> > adj(100001);
bool vis[100001] = {};
long long dm;
long long fm;
long long z = 0;
bool od;
void dfs(long long i, long long curd) {
  vis[i] = true;
  if (dm < curd) {
    dm = curd;
    fm = i;
  }
  long long cnt = 0;
  for (auto x : adj[i]) {
    if (adj[x].size() == 1) cnt++;
    if (!vis[x]) {
      dfs(x, curd + 1);
    }
  }
  if (curd % 2 and adj[i].size() == 1) od = 1;
  z += max(0LL, cnt - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  dm = 0;
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  long long mn = 1, mx = n - 1;
  long long strt;
  for (long long i = 1; i < n; i++) {
    if (adj[i].size() == 1) {
      strt = i;
      break;
    }
  }
  dfs(strt, 0);
  dm = inf;
  memset(vis, 0, sizeof(vis));
  od = false;
  z = 0;
  dfs(fm, 0);
  if (od) {
    mn = 3;
  }
  mx -= z;
  cout << mn << " " << mx;
}

### 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;
const long long mod = 1e9 + 7;
const long double pi = 3.1415926535897932384626433;
const long long inf = 1e18;
long long power(long long x, long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = ((res % mod) * (x % mod)) % mod;
    y = y >> 1;
    x = ((x % mod) * (x % mod)) % mod;
  }
  return res;
}
long long max(long long a, long long b) {
  long long ans = a > b ? a : b;
  return ans;
}
vector<vector<long long> > adj(100001);
bool vis[100001] = {};
long long dm;
long long fm;
long long z = 0;
bool od;
void dfs(long long i, long long curd) {
  vis[i] = true;
  if (dm < curd) {
    dm = curd;
    fm = i;
  }
  long long cnt = 0;
  for (auto x : adj[i]) {
    if (adj[x].size() == 1) cnt++;
    if (!vis[x]) {
      dfs(x, curd + 1);
    }
  }
  if (curd % 2 and adj[i].size() == 1) od = 1;
  z += max(0LL, cnt - 1);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n;
  cin >> n;
  dm = 0;
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  long long mn = 1, mx = n - 1;
  long long strt;
  for (long long i = 1; i < n; i++) {
    if (adj[i].size() == 1) {
      strt = i;
      break;
    }
  }
  dfs(strt, 0);
  dm = inf;
  memset(vis, 0, sizeof(vis));
  od = false;
  z = 0;
  dfs(fm, 0);
  if (od) {
    mn = 3;
  }
  mx -= z;
  cout << mn << " " << mx;
}

```