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: Runtime Error)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String {
            bytes
                .by_ref()
                .map(|r| r.unwrap() as char)
                .skip_while(|c| c.is_whitespace())
                .take_while(|c| !c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };

    ($next:expr, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
        }
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_imports)]
use std::cmp::{min, max};

fn main() {
    input!{
        s: chars
    }
    let n = s.len();
    let l = cycle_len(&s);
    if l == 1 {
        println!("{}", n);
        println!("1");
    } else if l == n {
        println!("1");
        println!("1");
    } else if l * 2 == n {
        println!("2");
        println!("{}", n - 1);
    } else {
        println!("2");
        println!("{}", n - n / l);
    }
}

fn cycle_len(s: &[char]) -> usize {
    let n = s.len();
    let mut div = divisors(n);
    div.sort();
    for &d in &div {
        for i in d..n {
            if s[i % d] != s[i] {
                break;
            }
        }
        return d;
    }
    n
}

fn factor(mut n: usize) -> Vec<(usize, usize)> {
    let mut ret = vec![];
    let mut p = 0;
    while n % 2 == 0 {
        p += 1;
        n /= 2;
    }
    if p != 0 {
        ret.push((2, p));
    }
    let mut r = 3;
    while r * r <= n {
        p = 0;
        while n % r == 0 {
            p += 1;
            n /= r;
        }
        if p != 0 {
            ret.push((r, p));
        }
        r += 2;
    }
    if n != 1 {
        ret.push((n, 1));
    }
    ret
}

fn divisors(n: usize) -> Vec<usize> {
    let mut ret = vec![];
    let fac = factor(n);
    let m = fac.len();
    let mut st = vec![0; m];
    while st[m - 1] <= fac[m - 1].1 {
        let mut d = 1;
        for i in 0..m {
            d *= fac[i].0.pow(st[i] as u32);
        }
        ret.push(d);
        st[0] += 1;
        let mut j = 0;
        while j < m - 1 && st[j] > fac[j].1 {
            st[j] = 0;
            st[j + 1] += 1;
            j += 1;
        }
    }
    ret
}