#include <bits/stdc++.h>
using namespace std;
inline long long int max(long long int a, long long int b) {
  return a > b ? a : b;
}
inline long long int min(long long int a, long long int b) {
  return a < b ? a : b;
}
const long long inf = 1e15;
const long long int MOD = 998244353;
void show(vector<long long int> v) {
  for (long long int i = 0; i < (v.size()); ++i) cout << v[i] << " ";
  cout << "\n";
}
void show(vector<int> v) {
  for (long long int i = 0; i < (v.size()); ++i) cout << v[i] << " ";
  cout << "\n";
}
void show(int a[], int arraysize) {
  for (int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(long long int a[], long long int arraysize) {
  for (long long int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(bool a[], long long int arraysize) {
  for (long long int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(set<int> s) {
  for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it << " ";
  }
  cout << "\n";
}
void show(map<long long int, long long int> mp) {
  for (auto it = mp.begin(); it != mp.end(); it++) {
    cout << it->first << " " << it->second << "\n";
  }
  cout << "\n";
}
long long int modpow(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = ((res % MOD) * (x % MOD)) % MOD;
    x = ((x % MOD) * (x % MOD)) % MOD;
    n >>= 1;
  }
  return res;
}
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x;
    x = x * x;
    n >>= 1;
  }
  return res;
}
const int N = 1e5 + 5;
vector<int> gp[N];
long long int r, mn, mx, e, l, m;
vector<int> leaf(N, 0);
bool u[N];
vector<int> d;
void dfs(long long int s, long long int t, long long int p) {
  if (gp[s].size() == 1) {
    l++;
    d.push_back(t % 2);
    leaf[s] = 1;
  }
  for (int i : gp[s]) {
    if (i != p) {
      dfs(i, t + 1, s);
    }
  }
  for (int i : gp[s]) {
    if (leaf[i] && i != p) {
      u[s] = 1;
      break;
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  e = n - 1;
  l = m = 0;
  for (long long int i = 0; i < (n - 1); ++i) {
    long long int a, b;
    cin >> a >> b;
    gp[a].push_back(b);
    gp[b].push_back(a);
  }
  memset(u, 0, sizeof(u));
  for (long long int i = 1; i < (n + 1); ++i) {
    if (gp[i].size() == 1) {
      u[gp[i][0]] = 1;
      r = i;
      break;
    }
  }
  dfs(r, 0, -1);
  mn = 1;
  sort(d.begin(), d.end());
  if (d[0] != d.back()) {
    mn = 3;
  }
  for (long long int i = 1; i < (n + 1); ++i) {
    if (u[i]) m++;
  }
  mx = e - l + m;
  cout << mn << " " << mx << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int t;
  t = 1;
  while (t) {
    solve();
    t--;
  }
  return 0;
}

### 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;
inline long long int max(long long int a, long long int b) {
  return a > b ? a : b;
}
inline long long int min(long long int a, long long int b) {
  return a < b ? a : b;
}
const long long inf = 1e15;
const long long int MOD = 998244353;
void show(vector<long long int> v) {
  for (long long int i = 0; i < (v.size()); ++i) cout << v[i] << " ";
  cout << "\n";
}
void show(vector<int> v) {
  for (long long int i = 0; i < (v.size()); ++i) cout << v[i] << " ";
  cout << "\n";
}
void show(int a[], int arraysize) {
  for (int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(long long int a[], long long int arraysize) {
  for (long long int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(bool a[], long long int arraysize) {
  for (long long int i = 0; i < arraysize; ++i) cout << a[i] << " ";
  cout << "\n";
}
void show(set<int> s) {
  for (auto it = s.begin(); it != s.end(); it++) {
    cout << *it << " ";
  }
  cout << "\n";
}
void show(map<long long int, long long int> mp) {
  for (auto it = mp.begin(); it != mp.end(); it++) {
    cout << it->first << " " << it->second << "\n";
  }
  cout << "\n";
}
long long int modpow(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = ((res % MOD) * (x % MOD)) % MOD;
    x = ((x % MOD) * (x % MOD)) % MOD;
    n >>= 1;
  }
  return res;
}
long long int power(long long int x, long long int n) {
  long long int res = 1;
  while (n > 0) {
    if (n & 1) res = res * x;
    x = x * x;
    n >>= 1;
  }
  return res;
}
const int N = 1e5 + 5;
vector<int> gp[N];
long long int r, mn, mx, e, l, m;
vector<int> leaf(N, 0);
bool u[N];
vector<int> d;
void dfs(long long int s, long long int t, long long int p) {
  if (gp[s].size() == 1) {
    l++;
    d.push_back(t % 2);
    leaf[s] = 1;
  }
  for (int i : gp[s]) {
    if (i != p) {
      dfs(i, t + 1, s);
    }
  }
  for (int i : gp[s]) {
    if (leaf[i] && i != p) {
      u[s] = 1;
      break;
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  e = n - 1;
  l = m = 0;
  for (long long int i = 0; i < (n - 1); ++i) {
    long long int a, b;
    cin >> a >> b;
    gp[a].push_back(b);
    gp[b].push_back(a);
  }
  memset(u, 0, sizeof(u));
  for (long long int i = 1; i < (n + 1); ++i) {
    if (gp[i].size() == 1) {
      u[gp[i][0]] = 1;
      r = i;
      break;
    }
  }
  dfs(r, 0, -1);
  mn = 1;
  sort(d.begin(), d.end());
  if (d[0] != d.back()) {
    mn = 3;
  }
  for (long long int i = 1; i < (n + 1); ++i) {
    if (u[i]) m++;
  }
  mx = e - l + m;
  cout << mn << " " << mx << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed << setprecision(12);
  int t;
  t = 1;
  while (t) {
    solve();
    t--;
  }
  return 0;
}

```