#include <bits/stdc++.h>
using namespace std;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ti;
vector<long long int> p[300005], f(300005, 0), h(300005, 0), d(300005, 0);
map<pair<long long int, long long int>, long long int> mp;
set<pair<long long int, long long int> > st;
set<long long int> s;
map<long long int, long long int> np, v;
queue<long long int> q;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res % 998244353;
}
long long int fact(long long int i) {
  if (i <= 0)
    return 1;
  else
    return (i * fact(i - 1)) % 998244353;
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
bool cmp(long long int i, long long int j) { return p[i].size() < p[j].size(); }
long long int l = 0, m = 0;
void dfs(long long int i, long long int pr) {
  d[i] = d[pr] + 1;
  long long int f = 0;
  for (auto u : p[i]) {
    if (u != pr) {
      dfs(u, i);
      if (p[u].size() == 1) {
        f = 1;
      }
    }
  }
  if (f) m++;
  if (p[i].size() == 1) {
    l++;
    s.insert(d[i] % 2);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int x, y, mix = 0, z;
  for (long long int i = 1; i < n; i++) {
    cin >> x >> y;
    p[x].push_back(y);
    p[y].push_back(x);
    if (p[x].size() > mix) {
      mix = p[x].size();
      z = x;
    }
    if (p[y].size() > mix) {
      mix = p[y].size();
      z = y;
    }
  }
  dfs(z, 0);
  long long int mi = 1, mx;
  if (s.size() == 2) {
    mi = 3;
  }
  mx = n - 1 - l + m;
  cout << mi << " " << mx << "\n";
}

### Prompt

Please formulate 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;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ti;
vector<long long int> p[300005], f(300005, 0), h(300005, 0), d(300005, 0);
map<pair<long long int, long long int>, long long int> mp;
set<pair<long long int, long long int> > st;
set<long long int> s;
map<long long int, long long int> np, v;
queue<long long int> q;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res % 998244353;
}
long long int fact(long long int i) {
  if (i <= 0)
    return 1;
  else
    return (i * fact(i - 1)) % 998244353;
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
bool cmp(long long int i, long long int j) { return p[i].size() < p[j].size(); }
long long int l = 0, m = 0;
void dfs(long long int i, long long int pr) {
  d[i] = d[pr] + 1;
  long long int f = 0;
  for (auto u : p[i]) {
    if (u != pr) {
      dfs(u, i);
      if (p[u].size() == 1) {
        f = 1;
      }
    }
  }
  if (f) m++;
  if (p[i].size() == 1) {
    l++;
    s.insert(d[i] % 2);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int x, y, mix = 0, z;
  for (long long int i = 1; i < n; i++) {
    cin >> x >> y;
    p[x].push_back(y);
    p[y].push_back(x);
    if (p[x].size() > mix) {
      mix = p[x].size();
      z = x;
    }
    if (p[y].size() > mix) {
      mix = p[y].size();
      z = y;
    }
  }
  dfs(z, 0);
  long long int mi = 1, mx;
  if (s.size() == 2) {
    mi = 3;
  }
  mx = n - 1 - l + m;
  cout << mi << " " << mx << "\n";
}

```