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`).

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.

Examples

Input

aab


Output

1
1


Input

bcbc


Output

2
3


Input

ddd


Output

3
1
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

std::vector<int> kmp(int n, char *s) {
  std::vector<int> fail(n, -1);
  for (int i = 1, j = -1; i < n; ++i) {
    while (j >= 0 && s[j + 1] != s[i]) j = fail[j];
    fail[i] = s[j + 1] == s[i] ? ++j : j;
  }
  return fail;
}

const int N = 5e5 + 10;

char s[N];

bool valid(int n, int b) {
  int p = n - b;
  return p == n || n % p;
}

int main() {
  scanf("%s", s);
  int n = strlen(s);
  auto u = kmp(n, s);
  if (valid(n, u[n - 1] + 1)) {
    puts("1\n1");
    return 0;
  }
  if (u[n - 1] == n - 2) {
    printf("%d\n1\n", n);
    return 0;
  }
  std::reverse(s, s + n);
  auto v = kmp(n, s);
  int ret = 0;
  for (int i = 0; i + 1 < n; ++i) {
    ret += valid(i + 1, u[i] + 1) && valid(n - 1 - i, v[n - 1 - i - 1] + 1);
  }
  printf("2\n%d\n", ret);
  return 0;
}
