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;
template <typename T>
inline void read(T &x) {
  static char _c;
  static bool _f;
  x = 0;
  _f = 0;
  _c = getchar();
  while (_c < '0' || '9' < _c) {
    if (_c == '-') _f = true;
    _c = getchar();
  }
  while ('0' <= _c && _c <= '9') {
    x = (x << 1) + (x << 3) + (_c & 15);
    _c = getchar();
  }
  if (_f) x = -x;
}
template <typename T>
inline void Min(T &x, T y) {
  if (y < x) x = y;
}
template <typename T>
inline void Max(T &x, T y) {
  if (x < y) x = y;
}
const int INF = 0x3f3f3f3f;
const double pi = (double)acos(-1.0);
const double eps = (double)1e-8;
int MOD = 998244353;
const int e5 = (int)1e5 + 5;
const int maxn = (int)1e5 + 20;
const int maxm = (int)1e6 + 20;
inline int sig(double x) { return x < -eps ? -1 : eps < x; }
long long fp(long long a, long long n, long long mod = MOD) {
  if (n < 0) a = fp(a, mod - 2, mod), n = -n;
  long long res = 1;
  for (; n; n >>= 1, a = a * a % mod)
    if (n & 1) res = res * a % mod;
  return res;
}
struct Mint {
  long long x;
  Mint() { x = 0; }
  Mint(int _x) : x(_x) {}
  Mint(long long _x) : x(_x) {}
  Mint operator-() const { return Mint(MOD - x); }
  Mint operator+(const Mint &rhs) const {
    long long res = x + rhs.x;
    if (res >= MOD) res -= MOD;
    return Mint(res);
  }
  Mint operator-(const Mint &rhs) const {
    long long res = x - rhs.x;
    if (res < 0) res += MOD;
    return Mint(res);
  }
  Mint operator*(const Mint &rhs) const { return Mint(x * rhs.x % MOD); }
  Mint operator/(const Mint &rhs) const {
    return Mint(x * fp(rhs.x, -1) % MOD);
  }
  Mint &operator+=(const Mint &rhs) {
    x += rhs.x;
    if (x >= MOD) x -= MOD;
    return *this;
  }
  Mint &operator*=(const Mint &rhs) {
    x = (x * rhs.x) % MOD;
    return *this;
  }
  bool operator==(const Mint &rhs) const { return x == rhs.x; }
  bool operator!=(const Mint &rhs) const { return x != rhs.x; }
  friend ostream &operator<<(ostream &out, const Mint &rhs) {
    return out << rhs.x;
  }
  friend istream &operator>>(istream &in, Mint &rhs) { return in >> rhs.x; }
};
int n;
vector<int> G[3 * e5];
int dep[3 * e5];
void dfs1(int u, int f) {
  dep[u] = dep[f] + 1;
  for (auto v : G[u]) {
    if (v == f) continue;
    dfs1(v, u);
  }
}
void dfs2(int u, int f, int &ans) {
  int vis = 0;
  for (auto v : G[u]) {
    if (v == f) continue;
    if (G[v].size() == 1) {
      ans += !vis;
      vis = 1;
    } else {
      ans++;
      dfs2(v, u, ans);
    }
  }
}
void work() {
  read(n);
  for (int i = 1; i < n; i++) {
    int x, y;
    read(x), read(y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  int rt = 1;
  dfs1(rt, 0);
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    if (dep[i] > mx) {
      mx = dep[i];
      rt = i;
    }
  }
  dfs1(rt, 0);
  int ans1 = 1;
  for (int i = 1; i <= n; i++) {
    if (G[i].size() == 1) {
      if (dep[i] % 2 == 0) {
        ans1 = 3;
        break;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (G[i].size() >= 2) {
      rt = i;
      break;
    }
  }
  int ans2 = 0;
  dfs2(rt, 0, ans2);
  cout << ans1 << " " << ans2 << endl;
}
int main(int argc, char **argv) {
  int tc = 1;
  for (int ca = 1; ca <= tc; ca++) {
    work();
  }
  return 0;
}
