Problem:
Score : 900 points

Problem Statement

Let x be a string of length at least 1.
We will call x a good string, if for any string y and any integer k (k \geq 2), the string obtained by concatenating k copies of y is different from x.
For example, a, bbc and cdcdc are good strings, while aa, bbbb and cdcdcd are not.

Let w be a string of length at least 1.
For a sequence F=(\,f_1,\,f_2,\,...,\,f_m) consisting of m elements,
we will call F a good representation of w, if the following conditions are both satisfied:

For any i \, (1 \leq i \leq m), f_i is a good string.

The string obtained by concatenating f_1,\,f_2,\,...,\,f_m in this order, is w.

For example, when w=aabb, there are five good representations of w:

(aabb)

(a,abb)

(aab,b)

(a,ab,b)

(a,a,b,b)

Among the good representations of w, the ones with the smallest number of elements are called the best representations of w.
For example, there are only one best representation of w=aabb: (aabb).

You are given a string w. Find the following:

the number of elements of a best representation of w

the number of the best representations of w, modulo 1000000007 \, (=10^9+7)

(It is guaranteed that a good representation of w always exists.)

Constraints

1 \leq |w| \leq 500000 \, (=5 \times 10^5)

w consists of lowercase letters (a-z).

Partial Score

400 points will be awarded for passing the test set satisfying 1 \leq |w| \leq 4000.

Input

The input is given from Standard Input in the following format:

w

Output

Print 2 lines.

In the first line, print the number of elements of a best representation of w.

In the second line, print the number of the best representations of w, modulo 1000000007.

Sample Input 1

aab

Sample Output 1

1
1

Sample Input 2

bcbc

Sample Output 2

2
3

In this case, there are 3 best representations of 2 elements:

(b,cbc)

(bc,bc)

(bcb,c)

Sample Input 3

ddd

Sample Output 3

3
1

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

int main() {
  string s;
  cin >> s;
  if (all_of(s.begin(), s.end(), [&](char c) { return c == s[0]; })) {
    cout << s.size() << "\n";
    cout << "1\n";
    return 0;
  }
  int n = s.size();
  constexpr int kP = 1'000'000'000 + 123;
  constexpr int kQ = 7122;
  vector<int> hv(n), pw(n + 1, 1);
  for (int i = 1; i <= n; ++i) pw[i] = 1LL * pw[i - 1] * kQ % kP;
  hv[0] = s[0];
  for (int i = 1; i < n; ++i) hv[i] = (1LL * hv[i - 1] * kQ + s[i]) % kP;

  vector<set<int>> is_cyclic(n);

  auto GetHash = [&](int p, int k) {
    if (p == 0) return hv[p + k - 1];
    return (int)((hv[p + k - 1] + kP - 1LL * pw[k] * hv[p - 1] % kP) % kP);
  };

  for (int k = 1; k <= n; ++k) {
    int h = hv[k - 1]; 
    for (int j = k; j <= n; j += k) {
      is_cyclic[j - 1].insert(k);
      if (j + k <= n && GetHash(j, k) != h) break;
    }
  }

  if (is_cyclic[n - 1].size() == 1) {
    cout << "1\n";
    cout << "1\n";
    return 0;
  }

  vector<set<int>> suffix(n);
  for (int k = 1; k <= n; ++k) {
    int h = GetHash(n - k, k);
    for (int j = n - k; j >= 0; j -= k) {
      suffix[j].insert(k);
      if (j >= k && GetHash(j - k, k) != h) break;
    }
  }
  int ans = 0;
  for (int i = 0; i + 1 < n; ++i) {
    if (is_cyclic[i].size() == 1 && suffix[i + 1].size() == 1) ans++;
  }
  assert(ans > 0);
  cout << "2\n";
  cout << ans << "\n";
  return 0;
}