#include <bits/stdc++.h>
using namespace std;
template <typename T, size_t N>
int SIZE(const T (&t)[N]) {
  return N;
}
template <typename T>
int SIZE(const T &t) {
  return t.size();
}
string to_string(const string s, int x1 = 0, int x2 = 1e9) {
  return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char c) { return string({c}); }
template <size_t N>
string to_string(const bitset<N> &b, int x1 = 0, int x2 = 1e9) {
  string t = "";
  for (int __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
       __iii__ <= __jjj__; ++__iii__) {
    t += b[__iii__] + '0';
  }
  return '"' + t + '"';
}
template <typename A, typename... C>
string to_string(const A(&v), int x1 = 0, int x2 = 1e9, C... coords);
int l_v_l_v_l = 0, t_a_b_s = 0;
template <typename A, typename B>
string to_string(const pair<A, B> &p) {
  l_v_l_v_l++;
  string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  l_v_l_v_l--;
  return res;
}
template <typename A, typename... C>
string to_string(const A(&v), int x1, int x2, C... coords) {
  int rnk = rank<A>::value;
  string tab(t_a_b_s, ' ');
  string res = "";
  bool first = true;
  if (l_v_l_v_l == 0) res += '\n';
  res += tab + "[";
  x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
  auto l = begin(v);
  advance(l, x1);
  auto r = l;
  advance(r, (x2 - x1) + (x2 < SIZE(v)));
  for (auto e = l; e != r; e = next(e)) {
    if (!first) {
      res += ", ";
    }
    first = false;
    l_v_l_v_l++;
    if (e != l) {
      if (rnk > 1) {
        res += '\n';
        t_a_b_s = l_v_l_v_l;
      };
    } else {
      t_a_b_s = 0;
    }
    res += to_string(*e, coords...);
    l_v_l_v_l--;
  }
  res += "]";
  if (l_v_l_v_l == 0) res += '\n';
  return res;
}
void dbgm() { ; }
template <typename Heads, typename... Tails>
void dbgm(Heads H, Tails... T) {
  cerr << to_string(H) << " | ";
  dbgm(T...);
}
bool vectorpair_sort_by_sec(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second < b.second;
}
bool custom_compare(const pair<int, int> &p1, const pair<int, int> &p2) {
  if (p1.second < p2.second)
    return true;
  else
    return false;
}
long long _pow(long long a, long long b) {
  long long ans = 1;
  for (int i = 1; i <= b; i++) {
    ans = (ans * a) % 998244353;
  }
  return ans;
}
void prep() {}
void testcase() {
  long long n;
  cin >> n;
  long long m = 0, u = 0;
  while (1) {
    bool ok = false;
    for (long long i = 1; i < n; i++) {
      long long k = n * i;
      long long a = (long long)sqrt(k);
      if (a * a == k) {
        m++;
        n = sqrt(k);
        if (i != 1) u++;
        ok = true;
      }
      if (ok) break;
    }
    if (ok)
      continue;
    else
      break;
  }
  if (u) m++;
  cout << n << " " << m;
}
int main() {
  time_t t1, t2;
  t1 = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  prep();
  testcase();
}

### Prompt

Please create a solution in CPP to the following problem:

JATC's math teacher always gives the class some interesting math problems so that they don't get bored. Today the problem is as follows. Given an integer n, you can perform the following operations zero or more times:

  * mul x: multiplies n by x (where x is an arbitrary positive integer). 
  * sqrt: replaces n with √{n} (to apply this operation, √{n} must be an integer). 



You can perform these operations as many times as you like. What is the minimum value of n, that can be achieved and what is the minimum number of operations, to achieve that minimum value?

Apparently, no one in the class knows the answer to this problem, maybe you can help them?

Input

The only line of the input contains a single integer n (1 ≤ n ≤ 10^6) — the initial number.

Output

Print two integers: the minimum integer n that can be achieved using the described operations and the minimum number of operations required.

Examples

Input

20


Output

10 2

Input

5184


Output

6 4

Note

In the first example, you can apply the operation mul 5 to get 100 and then sqrt to get 10.

In the second example, you can first apply sqrt to get 72, then mul 18 to get 1296 and finally two more sqrt and you get 6.

Note, that even if the initial value of n is less or equal 10^6, it can still become greater than 10^6 after applying one or more operations.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
template <typename T, size_t N>
int SIZE(const T (&t)[N]) {
  return N;
}
template <typename T>
int SIZE(const T &t) {
  return t.size();
}
string to_string(const string s, int x1 = 0, int x2 = 1e9) {
  return '"' + ((x1 < s.size()) ? s.substr(x1, x2 - x1 + 1) : "") + '"';
}
string to_string(const char *s) { return to_string((string)s); }
string to_string(const bool b) { return (b ? "true" : "false"); }
string to_string(const char c) { return string({c}); }
template <size_t N>
string to_string(const bitset<N> &b, int x1 = 0, int x2 = 1e9) {
  string t = "";
  for (int __iii__ = min(x1, SIZE(b)), __jjj__ = min(x2, SIZE(b) - 1);
       __iii__ <= __jjj__; ++__iii__) {
    t += b[__iii__] + '0';
  }
  return '"' + t + '"';
}
template <typename A, typename... C>
string to_string(const A(&v), int x1 = 0, int x2 = 1e9, C... coords);
int l_v_l_v_l = 0, t_a_b_s = 0;
template <typename A, typename B>
string to_string(const pair<A, B> &p) {
  l_v_l_v_l++;
  string res = "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  l_v_l_v_l--;
  return res;
}
template <typename A, typename... C>
string to_string(const A(&v), int x1, int x2, C... coords) {
  int rnk = rank<A>::value;
  string tab(t_a_b_s, ' ');
  string res = "";
  bool first = true;
  if (l_v_l_v_l == 0) res += '\n';
  res += tab + "[";
  x1 = min(x1, SIZE(v)), x2 = min(x2, SIZE(v));
  auto l = begin(v);
  advance(l, x1);
  auto r = l;
  advance(r, (x2 - x1) + (x2 < SIZE(v)));
  for (auto e = l; e != r; e = next(e)) {
    if (!first) {
      res += ", ";
    }
    first = false;
    l_v_l_v_l++;
    if (e != l) {
      if (rnk > 1) {
        res += '\n';
        t_a_b_s = l_v_l_v_l;
      };
    } else {
      t_a_b_s = 0;
    }
    res += to_string(*e, coords...);
    l_v_l_v_l--;
  }
  res += "]";
  if (l_v_l_v_l == 0) res += '\n';
  return res;
}
void dbgm() { ; }
template <typename Heads, typename... Tails>
void dbgm(Heads H, Tails... T) {
  cerr << to_string(H) << " | ";
  dbgm(T...);
}
bool vectorpair_sort_by_sec(const pair<int, int> &a, const pair<int, int> &b) {
  return a.second < b.second;
}
bool custom_compare(const pair<int, int> &p1, const pair<int, int> &p2) {
  if (p1.second < p2.second)
    return true;
  else
    return false;
}
long long _pow(long long a, long long b) {
  long long ans = 1;
  for (int i = 1; i <= b; i++) {
    ans = (ans * a) % 998244353;
  }
  return ans;
}
void prep() {}
void testcase() {
  long long n;
  cin >> n;
  long long m = 0, u = 0;
  while (1) {
    bool ok = false;
    for (long long i = 1; i < n; i++) {
      long long k = n * i;
      long long a = (long long)sqrt(k);
      if (a * a == k) {
        m++;
        n = sqrt(k);
        if (i != 1) u++;
        ok = true;
      }
      if (ok) break;
    }
    if (ok)
      continue;
    else
      break;
  }
  if (u) m++;
  cout << n << " " << m;
}
int main() {
  time_t t1, t2;
  t1 = clock();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ;
  prep();
  testcase();
}

```