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: Wrong Answer)

#![allow(unused_imports, unused_variables, dead_code)]
use std::io::*;
use std::fmt::*;
use std::str::*;
use std::cmp::*;
use std::collections::*;

trait InputValue {
    fn parse(s: &str) -> Self;
}

fn read<T: InputValue>() -> T {
    let mut buf = String::new();
    let _ = stdin().read_line(&mut buf);
    T::parse(&buf.trim())
}

fn readnc<T: InputValue>() -> Vec<T> {
    let mut vec = vec![];
    let line: String = read();
    for token in line.split_whitespace() {
        vec.push(T::parse(token));
    }
    vec
}

fn readn<T: InputValue>(n: usize) -> Vec<T> {
    let mut vec = vec![];
    for _ in 0..n {
        vec.push(read());
    }
    vec
}

macro_rules! parse_single_value {
    ($($t:ty),*) => {
        $(
            impl InputValue for $t {
                fn parse(s: &str) -> $t { s.parse().unwrap() }
            }
        )*
	}
}
parse_single_value!(i32, i64, f32, f64, usize, String);

macro_rules! parse_tuple {
	($($t:ident),*) => {
		impl<$($t),*> InputValue for ($($t),*) where $($t: InputValue),* {
			fn parse(s: &str) -> ($($t),*) {
				let mut tokens = s.split_whitespace();
				let t = ($($t::parse(tokens.next().unwrap())),*);
				t
			}
		}
	}
}
parse_tuple!(A, B);
parse_tuple!(A, B, C);

// ===

fn error_function(a: &Vec<u8>) -> Vec<usize> {
    let n = a.len();
    let mut err: Vec<usize> = vec![0; n+1];

    for i in 2..n+1 {
        let mut now = err[i-1];
        while now > 0 && a[i-1] != a[now] {
            now = err[now];
        }
        if a[i-1] == a[now] {
            err[i] = now + 1;
        } else {
            err[i] = 0;
        }
    }
    err
}

fn main() {
    let a: String = read();
    let a = a.into_bytes();

    let n = a.len();
    let prefix = error_function(&a);

    let mut ra = a;
    ra.reverse();
    let suffix = error_function(&ra);

    let cycle = n - prefix[n];
    if cycle == n {
        println!("{}\n{}", 1, 1);
    } else if cycle == 1 {
        println!("{}\n{}", n, 1);
    } else {
        let mut cnt = n - 1;
        if n / cycle >= 3 {
            cnt -= ((n / cycle) - 1);
        }
        for x in 2..cycle {
            if prefix[x] >= 1 && x % (x - prefix[x]) == 0 {
                cnt -= 1;
            }
            if suffix[x] >= 1 && x % (x - suffix[x]) == 0 {
                cnt -= 1;
            }
        }
        println!("{}\n{}", 2, cnt);
    }
}