#include <bits/stdc++.h>
using namespace std;
const double pi = 3.14159265;
const long long mod = 1e9 + 7;
const long long INF = 2e18;
const long long N = 1e5 + 5;
const long long P0 = 437;
const long long P1 = 343;
vector<long long> v[N], ch(N), nch(N), was(N);
bool fl = true;
void dfs(long long pos) {
  was[pos] = 1;
  long long c0 = 0, c1 = 0;
  bool q = false;
  for (long long x : v[pos]) {
    if (!was[x]) {
      q = true;
      dfs(x);
      if (nch[x]) c1 = 1;
      if (ch[x]) c0 = 1;
    }
  }
  if (q == false) {
    nch[pos] = 1;
    return;
  }
  if (c0 && c1) fl = false;
  if (c0) nch[pos] = 1;
  if (c1) ch[pos] = 1;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> cnt(n), L(n - 1), R(n - 1);
  for (long long i = 0; i < n - 1; i++) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    v[l].emplace_back(r);
    v[r].emplace_back(l);
    cnt[l]++;
    cnt[r]++;
    L[i] = l;
    R[i] = r;
  }
  vector<long long> cnt2(n);
  for (long long i = 0; i < n - 1; i++) {
    if (cnt[L[i]] == 1) cnt2[R[i]]++;
    if (cnt[R[i]] == 1) cnt2[L[i]]++;
  }
  long long ma = n - 1;
  for (long long i = 0; i < n; i++)
    if (cnt2[i] > 1) ma -= (cnt2[i] - 1);
  dfs(0);
  if (cnt[0] == 1 && ch[0] == 1)
    cout << 3 << ' ' << ma << '\n';
  else if (fl)
    cout << 1 << ' ' << ma << '\n';
  else
    cout << 3 << ' ' << ma << '\n';
}

### Prompt

Generate 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 double pi = 3.14159265;
const long long mod = 1e9 + 7;
const long long INF = 2e18;
const long long N = 1e5 + 5;
const long long P0 = 437;
const long long P1 = 343;
vector<long long> v[N], ch(N), nch(N), was(N);
bool fl = true;
void dfs(long long pos) {
  was[pos] = 1;
  long long c0 = 0, c1 = 0;
  bool q = false;
  for (long long x : v[pos]) {
    if (!was[x]) {
      q = true;
      dfs(x);
      if (nch[x]) c1 = 1;
      if (ch[x]) c0 = 1;
    }
  }
  if (q == false) {
    nch[pos] = 1;
    return;
  }
  if (c0 && c1) fl = false;
  if (c0) nch[pos] = 1;
  if (c1) ch[pos] = 1;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  vector<long long> cnt(n), L(n - 1), R(n - 1);
  for (long long i = 0; i < n - 1; i++) {
    long long l, r;
    cin >> l >> r;
    l--;
    r--;
    v[l].emplace_back(r);
    v[r].emplace_back(l);
    cnt[l]++;
    cnt[r]++;
    L[i] = l;
    R[i] = r;
  }
  vector<long long> cnt2(n);
  for (long long i = 0; i < n - 1; i++) {
    if (cnt[L[i]] == 1) cnt2[R[i]]++;
    if (cnt[R[i]] == 1) cnt2[L[i]]++;
  }
  long long ma = n - 1;
  for (long long i = 0; i < n; i++)
    if (cnt2[i] > 1) ma -= (cnt2[i] - 1);
  dfs(0);
  if (cnt[0] == 1 && ch[0] == 1)
    cout << 3 << ' ' << ma << '\n';
  else if (fl)
    cout << 1 << ' ' << ma << '\n';
  else
    cout << 3 << ' ' << ma << '\n';
}

```