#include <bits/stdc++.h>
const long long int MOD = 1e9 + 7;
const long long int INF = 1011111111;
const long long int LLINF = 1000111000111000111LL;
const long double EPS = 1e-10;
const long double PI = 3.14159265358979323;
using namespace std;
long long int power(long long int base, long long int exponent) {
  long long int ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base);
    base = (1LL * base * base);
    exponent >>= 1;
  }
  return ans;
}
const long long int N = 1e5 + 5;
vector<long long int> v[N];
bool visit[N];
long long int leaves = 0;
long long int level[N];
void dfs(long long int node, long long int x) {
  visit[node] = true;
  level[node] = x;
  for (auto to : v[node]) {
    if (!visit[to]) {
      dfs(to, x + 1);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j, k;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  vector<long long int> tt;
  set<long long int> xx;
  for (i = 1; i <= n; i++) {
    if (v[i].size() == 1) {
      leaves++;
      tt.push_back(i);
      xx.insert(v[i][0]);
    }
  }
  long long int maxi = (long long int)xx.size() + (n - leaves) - 1;
  dfs(tt[0], 1);
  vector<long long int> y;
  for (i = 0; i < (long long int)tt.size(); i++) {
    y.push_back(level[tt[i]]);
  }
  sort(y.begin(), y.end());
  bool ok = false;
  long long int mini;
  for (i = 0; i < (long long int)y.size(); i++)
    if (y[i] % 2 == 0) ok = true;
  if (ok)
    mini = 3;
  else
    mini = 1;
  cout << mini << " " << maxi;
  cerr << '\n'
       << "Time elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\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>
const long long int MOD = 1e9 + 7;
const long long int INF = 1011111111;
const long long int LLINF = 1000111000111000111LL;
const long double EPS = 1e-10;
const long double PI = 3.14159265358979323;
using namespace std;
long long int power(long long int base, long long int exponent) {
  long long int ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base);
    base = (1LL * base * base);
    exponent >>= 1;
  }
  return ans;
}
const long long int N = 1e5 + 5;
vector<long long int> v[N];
bool visit[N];
long long int leaves = 0;
long long int level[N];
void dfs(long long int node, long long int x) {
  visit[node] = true;
  level[node] = x;
  for (auto to : v[node]) {
    if (!visit[to]) {
      dfs(to, x + 1);
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n, m, i, j, k;
  cin >> n;
  for (i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  vector<long long int> tt;
  set<long long int> xx;
  for (i = 1; i <= n; i++) {
    if (v[i].size() == 1) {
      leaves++;
      tt.push_back(i);
      xx.insert(v[i][0]);
    }
  }
  long long int maxi = (long long int)xx.size() + (n - leaves) - 1;
  dfs(tt[0], 1);
  vector<long long int> y;
  for (i = 0; i < (long long int)tt.size(); i++) {
    y.push_back(level[tt[i]]);
  }
  sort(y.begin(), y.end());
  bool ok = false;
  long long int mini;
  for (i = 0; i < (long long int)y.size(); i++)
    if (y[i] % 2 == 0) ok = true;
  if (ok)
    mini = 3;
  else
    mini = 1;
  cout << mini << " " << maxi;
  cerr << '\n'
       << "Time elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms"
       << '\n';
}

```