#include <bits/stdc++.h>
using namespace std;
long long const mod = 998244353;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return (((a % mod) * (b % mod)) % mod + mod) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modInv(long long x) { return power(x, mod - 2, mod) % mod; }
bool first = false;
bool notuse[100001];
void func1(long long s, vector<long long> v[], long long p, long long d) {
  if (v[s].size() == 1) {
    if (p != -1) {
      if (d % 2 != 0) {
        first = true;
      }
      return;
    }
  }
  for (auto x : v[s]) {
    if (x != p) func1(x, v, s, d + 1);
  }
}
long long func2(long long s, vector<long long> v[], long long p) {
  long long ans = 0;
  if (notuse[s]) return 0;
  for (auto x : v[s]) {
    if (x != p && notuse[x] == false) ans = ans + 1 + func2(x, v, s);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> v[n + 1];
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  long long id = 1;
  for (long long i = 1; i <= n; i++) {
    if (v[i].size() == 1) {
      id = i;
      break;
    }
  }
  func1(id, v, -1, 0);
  long long mn;
  if (first)
    mn = 3;
  else {
    mn = 1;
  }
  cout << mn << " ";
  long long mx = 0;
  memset(notuse, false, sizeof(notuse));
  for (long long i = 1; i <= n; i++) {
    vector<long long> p;
    for (auto x : v[i]) {
      if (v[x].size() == 1) {
        p.push_back(x);
      }
    }
    long long pz = p.size();
    if (pz > 1) {
      for (long long j = 1; j < pz; j++) notuse[p[j]] = true;
    }
  }
  for (long long i = 1; i <= n; i++)
    if (notuse[i] == false) {
      id = i;
      break;
    }
  cout << func2(id, v, -1) << endl;
  return 0;
}

### 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;
long long const mod = 998244353;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return (((a % mod) * (b % mod)) % mod + mod) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modInv(long long x) { return power(x, mod - 2, mod) % mod; }
bool first = false;
bool notuse[100001];
void func1(long long s, vector<long long> v[], long long p, long long d) {
  if (v[s].size() == 1) {
    if (p != -1) {
      if (d % 2 != 0) {
        first = true;
      }
      return;
    }
  }
  for (auto x : v[s]) {
    if (x != p) func1(x, v, s, d + 1);
  }
}
long long func2(long long s, vector<long long> v[], long long p) {
  long long ans = 0;
  if (notuse[s]) return 0;
  for (auto x : v[s]) {
    if (x != p && notuse[x] == false) ans = ans + 1 + func2(x, v, s);
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  vector<long long> v[n + 1];
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  long long id = 1;
  for (long long i = 1; i <= n; i++) {
    if (v[i].size() == 1) {
      id = i;
      break;
    }
  }
  func1(id, v, -1, 0);
  long long mn;
  if (first)
    mn = 3;
  else {
    mn = 1;
  }
  cout << mn << " ";
  long long mx = 0;
  memset(notuse, false, sizeof(notuse));
  for (long long i = 1; i <= n; i++) {
    vector<long long> p;
    for (auto x : v[i]) {
      if (v[x].size() == 1) {
        p.push_back(x);
      }
    }
    long long pz = p.size();
    if (pz > 1) {
      for (long long j = 1; j < pz; j++) notuse[p[j]] = true;
    }
  }
  for (long long i = 1; i <= n; i++)
    if (notuse[i] == false) {
      id = i;
      break;
    }
  cout << func2(id, v, -1) << endl;
  return 0;
}

```