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>
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 119 << 23 | 1;
const int INF = (int)1e9 + 23111992;
const long long LINF = (long long)1e18 + 23111992;
const long double PI = acos((long double)-1);
const long double EPS = 1e-6;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline unsigned long long isqrt(unsigned long long k) {
  unsigned long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
mt19937 mt(chrono::high_resolution_clock::now().time_since_epoch().count());
inline int mrand() { return abs((int)mt()); }
inline int mrand(int k) { return abs((int)mt()) % k; }
void chemthan() {
  int n;
  cin >> n;
  vector<vector<int> > adj(n);
  for (int i = (0); i < (n - 1); ++i) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v), adj[v].push_back(u);
  }
  vector<int> lev(n);
  function<void(int, int)> dfs = [&](int u, int p) {
    for (int v : adj[u]) {
      if (v ^ p) {
        lev[v] = lev[u] ^ 1;
        dfs(v, u);
      }
    }
  };
  for (int u = (0); u < (n); ++u)
    if (int((adj[u]).size()) == 1) {
      dfs(u, -1);
      break;
    }
  int mn = 1;
  for (int i = (0); i < (n); ++i)
    if (int((adj[i]).size()) == 1 && (lev[i] & 1)) mn = 3;
  int mx = n - 1;
  vector<int> dj(n);
  for (int i = (0); i < (n); ++i) dj[i] = i;
  function<int(int)> find = [&](int u) {
    return dj[u] == u ? dj[u] : dj[u] = find(dj[u]);
  };
  for (int u = (0); u < (n); ++u) {
    vector<int> leaves;
    for (int v : adj[u]) {
      if (int((adj[v]).size()) == 1) {
        leaves.push_back(v);
      }
    }
    for (int i = (1); i < (int((leaves).size())); ++i) {
      int x = leaves[0];
      int y = leaves[i];
      dj[find(x)] = find(y);
    }
  }
  map<int, int> hs;
  for (int i = (0); i < (n); ++i) hs[find(i)]++;
  for (auto e : hs) {
    mx -= e.second - 1;
  }
  cout << mn << " " << mx << "\n";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  chemthan();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
