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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::BufRead;

fn main() {
    let stdin = std::io::stdin();
    let mut w = stdin.lock().gets().into_bytes();
    let mut badsp = vec![false; w.len() + 1];
    check_from_left(&w, &mut badsp);
    w.reverse();
    badsp.reverse();
    check_from_left(&w, &mut badsp);
    if !badsp[0] && !badsp[w.len()] {
        println!("1");
        println!("1");
    } else {
        let count = badsp[1..w.len()].into_iter().filter(|&x| !x).count();
        if count == 0 {
            println!("{}", w.len());
            println!("1");
        } else {
            println!("2");
            println!("{}", count);
        }
    }
}

fn check_from_left(w: &[u8], badsp: &mut [bool]) {
    let mut done = vec![false; w.len() / 2 + 1];
    for len in 1..w.len() / 2 + 1 {
        if done[len] {
            continue;
        }
        let mut start = len;
        while start + len <= w.len() && w[..len] == w[start..start + len] {
            badsp[start + len] = true;
            start += len;
        }
        let mut mul = len * 2;
        while mul <= std::cmp::min(w.len() / 2, start) {
            done[mul] = true;
            mul += len;
        }
    }
}

trait Gets { fn gets(&mut self) -> String; }
impl<T: std::io::BufRead> Gets for T {
    fn gets(&mut self) -> String { self.lines().next().unwrap().unwrap() } }