#include <bits/stdc++.h>
using namespace std;
long long int binpowmod(long long int a, long long int b, long long int m) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return binpowmod(n, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, m, n, i, f = 0, j, ans = 0, u, v, s;
  cin >> n;
  vector<vector<long long int>> adj(n + 1);
  vector<long long int> used(n + 1, 0);
  for (i = 0; i < n - 1; ++i) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
    used[u]++;
    used[v]++;
  }
  for (i = 1; i < n + 1; ++i) {
    if (used[i] == 1) {
      s = i;
      break;
    }
  }
  vector<long long int> d(n + 1, -1), p(n + 1, -1);
  vector<long long int> lf;
  queue<long long int> q;
  q.push(s);
  d[s] = 0;
  lf.push_back(s);
  while (!q.empty()) {
    u = q.front();
    q.pop();
    t = 0;
    for (auto k : adj[u]) {
      if (d[k] == -1) {
        if (s == u) p[s] = k;
        t++;
        q.push(k);
        d[k] = d[u] + 1;
        p[k] = u;
      }
    }
    if (t == 0) lf.push_back(u);
  }
  long long int mmin, mmax = 0;
  long long int flag = 1;
  for (i = 0; i < lf.size(); ++i) {
    if (d[lf[i]] % 2 == 0)
      continue;
    else {
      flag = 0;
      break;
    }
  }
  if (flag == 1)
    mmin = 1;
  else
    mmin = 3;
  ans = n - 1;
  map<long long int, long long int> ma;
  for (i = 0; i < lf.size(); ++i) {
    ma[p[lf[i]]]++;
  }
  for (auto itr = ma.begin(); itr != ma.end(); ++itr) {
    long long int temp = itr->second;
    ans -= (temp - 1);
  }
  cout << mmin << " " << ans;
  return 0;
}

### 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;
long long int binpowmod(long long int a, long long int b, long long int m) {
  a %= m;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % m;
    a = a * a % m;
    b >>= 1;
  }
  return res;
}
long long int binpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int modInverse(long long int n, long long int p) {
  return binpowmod(n, p - 2, p);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t, m, n, i, f = 0, j, ans = 0, u, v, s;
  cin >> n;
  vector<vector<long long int>> adj(n + 1);
  vector<long long int> used(n + 1, 0);
  for (i = 0; i < n - 1; ++i) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
    used[u]++;
    used[v]++;
  }
  for (i = 1; i < n + 1; ++i) {
    if (used[i] == 1) {
      s = i;
      break;
    }
  }
  vector<long long int> d(n + 1, -1), p(n + 1, -1);
  vector<long long int> lf;
  queue<long long int> q;
  q.push(s);
  d[s] = 0;
  lf.push_back(s);
  while (!q.empty()) {
    u = q.front();
    q.pop();
    t = 0;
    for (auto k : adj[u]) {
      if (d[k] == -1) {
        if (s == u) p[s] = k;
        t++;
        q.push(k);
        d[k] = d[u] + 1;
        p[k] = u;
      }
    }
    if (t == 0) lf.push_back(u);
  }
  long long int mmin, mmax = 0;
  long long int flag = 1;
  for (i = 0; i < lf.size(); ++i) {
    if (d[lf[i]] % 2 == 0)
      continue;
    else {
      flag = 0;
      break;
    }
  }
  if (flag == 1)
    mmin = 1;
  else
    mmin = 3;
  ans = n - 1;
  map<long long int, long long int> ma;
  for (i = 0; i < lf.size(); ++i) {
    ma[p[lf[i]]]++;
  }
  for (auto itr = ma.begin(); itr != ma.end(); ++itr) {
    long long int temp = itr->second;
    ans -= (temp - 1);
  }
  cout << mmin << " " << ans;
  return 0;
}

```