Problem:
Score : 1400 points

Problem Statement

Consider the following set of rules for encoding strings consisting of 0 and 1:

Strings 0 and 1 can be encoded as 0 and 1, respectively.

If strings A and B can be encoded as P and Q, respectively, then string AB can be encoded as PQ.

If string A can be encoded as P and K \geq 2 is a positive integer, then string AA...A (A repeated K times) can be encoded as (PxK).

For example, string 001001001, among other possibilities, can be encoded as 001001001, 00(1(0x2)x2)1 and (001x3).

Let's call string A a subset of string B if:

A and B are equal in length and consist of 0 and 1;

for all indices i such that A_i = 1, it's also true that B_i = 1.

You are given string S consisting of 0 and 1. Find the total number of distinct encodings of all subsets of S, modulo 998244353.

Constraints

1 \leq |S| \leq 100

S consists of 0 and 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the total number of distinct encodings of all subsets of S modulo 998244353.

Sample Input 1

011

Sample Output 1

9

There are four subsets of S:

011 can be encoded as 011 and 0(1x2);

010 can be encoded as 010;

001 can be encoded as 001 and (0x2)1;

000 can be encoded as 000, 0(0x2), (0x2)0 and (0x3).

Thus, the total number of encodings of all subsets of S is 2 + 1 + 2 + 4 = 9.

Sample Input 2

0000

Sample Output 2

10

This time S has only one subset, but it can be encoded in 10 different ways.

Sample Input 3

101110

Sample Output 3

156

Sample Input 4

001110111010110001100000100111

Sample Output 4

363383189

Don't forget to take the result modulo 998244353.

Solution in Rust (1.15.1) (judge verdict: Accepted)

// ---------- begin ModInt ----------
const MOD: u32 = 998_244_353;

#[derive(Clone, Copy)]
struct ModInt(u32);

impl std::ops::Add for ModInt {
    type Output = ModInt;
    fn add(self, rhs: ModInt) -> Self::Output {
        let mut d = self.0 + rhs.0;
        if d >= MOD {
            d -= MOD;
        }
        ModInt(d)
    }
}

impl std::ops::AddAssign for ModInt {
    fn add_assign(&mut self, rhs: ModInt) {
        *self = *self + rhs;
    }
}

impl std::ops::Sub for ModInt {
    type Output = ModInt;
    fn sub(self, rhs: ModInt) -> Self::Output {
        let mut d = self.0 + MOD - rhs.0;
        if d >= MOD {
            d -= MOD;
        }
        ModInt(d)
    }
}

impl std::ops::SubAssign for ModInt {
    fn sub_assign(&mut self, rhs: ModInt) {
        *self = *self - rhs;
    }
}

impl std::ops::Mul for ModInt {
    type Output = ModInt;
    fn mul(self, rhs: ModInt) -> Self::Output {
        ModInt((self.0 as u64 * rhs.0 as u64 % MOD as u64) as u32)
    }
}

impl std::ops::MulAssign for ModInt {
    fn mul_assign(&mut self, rhs: ModInt) {
        *self = *self * rhs;
    }
}

impl std::ops::Neg for ModInt {
    type Output = ModInt;
    fn neg(self) -> Self::Output {
        ModInt(if self.0 == 0 {0} else {MOD - self.0})
    }
}

impl std::fmt::Display for ModInt {
    fn fmt<'a>(&self, f: &mut std::fmt::Formatter<'a>) -> std::fmt::Result {
        write!(f, "{}", self.0)
    }
}

impl std::str::FromStr for ModInt {
    type Err = std::num::ParseIntError;
    fn from_str(s: &str) -> Result<Self, Self::Err> {
        let val = s.parse::<u32>()?;
        Ok(ModInt::new(val))
    }
}

impl From<usize> for ModInt {
    fn from(val: usize) -> ModInt {
        ModInt((val % MOD as usize) as u32)
    }
}

#[allow(dead_code)]
impl ModInt {
    pub fn new(n: u32) -> ModInt {
        ModInt(n % MOD)
    }
    pub fn zero() -> ModInt {
        ModInt(0)
    }
    pub fn one() -> ModInt {
        ModInt(1)
    }
    pub fn pow(self, mut n: u32) -> ModInt {
        let mut t = ModInt::one();
        let mut s = self;
        while n > 0 {
            if n & 1 == 1 {
                t *= s;
            }
            s *= s;
            n >>= 1;
        }
        t
    }
    pub fn inv(self) -> ModInt {
        assert!(self.0 > 0);
        self.pow(MOD - 2)
    }
}
// ---------- end ModInt ----------
// ---------- begin Precalc ----------
#[allow(dead_code)]
struct Precalc {
    inv: Vec<ModInt>,
    fact: Vec<ModInt>,
    ifact: Vec<ModInt>,
}

#[allow(dead_code)]
impl Precalc {
    pub fn new(n: usize) -> Precalc {
        let mut inv = vec![ModInt::one(); n + 1];
        let mut fact = vec![ModInt::one(); n + 1];
        let mut ifact = vec![ModInt::one(); n + 1];
        for i in 2..(n + 1) {
            inv[i] = -inv[MOD as usize % i] * ModInt(MOD / i as u32);
            fact[i] = fact[i - 1] * ModInt(i as u32);
            ifact[i] = ifact[i - 1] * inv[i];
        }
        Precalc {
            inv: inv,
            fact: fact,
            ifact: ifact,
        }
    }
    pub fn inv(&self, n: usize) -> ModInt {
        self.inv[n]
    }
    pub fn fact(&self, n: usize) -> ModInt {
        self.fact[n]
    }
    pub fn ifact(&self, n: usize) -> ModInt {
        self.ifact[n]
    }
    pub fn comb(&self, n: usize, k: usize) -> ModInt {
        if k > n {
            return ModInt::zero();
        }
        self.fact[n] * self.ifact[k] * self.ifact[n - k]
    }
}
// ---------- end Precalc ----------

// サブセットを無視して一つの文字列について考える
// f(S[0..n]) = f(S[1..n]) + (長さxをk回繰り返すことが可能なx, kについて) f[0..x] * f[(k * x)..n]
// こんな感じのDPで解ける
// サブセットありだと？
// 求めたいものをgとおく
// g(S[0..n])への寄与を考える
// 先頭が0だと1, 1だと2の係数でg(S[1..n])
// 長さxをk回繰り返すyatuは長さxのbitwise and をk個とった文字列Tについて
// g(T[0..x]) * g(S[(k * x)..n])
//
// Tの数がどれくらいで抑えられるの
// 少ないと信じれば解けそうだが

fn calc(s: &[u8], memo: &mut std::collections::BTreeMap<Vec<u8>, ModInt>) -> ModInt {
    if s.len() == 0 {
        return ModInt::one();
    }
    let s: Vec<u8> = s.iter().cloned().collect();
    if let Some(&v) = memo.get(&s) {
        return v;
    }
    let mut ans = calc(&s[1..], memo);
    if s[0] == 1 {
        ans *= ModInt(2);
    }
    let n = s.len();
    for len in 1..(n + 1) {
        if 2 * len > n {
            break;
        }
        let mut x: Vec<_> = s[..len].iter().cloned().collect();
        for k in 2.. {
            if k * len > n {
                break;
            }
            for (x, d) in x.iter_mut().zip(s.iter().skip((k - 1) * len)) {
                *x &= *d;
            }
            let a = calc(&x, memo);
            let b = calc(&s[(k * len)..], memo);
            ans += a * b;
        }
    }
    memo.insert(s, ans);
    ans
}

fn run() {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let s: Vec<u8> = s.trim().chars().map(|c| c.to_digit(10).unwrap() as u8).collect();
    let mut memo = std::collections::BTreeMap::new();
    let ans = calc(&s, &mut memo);
    println!("{}", ans);
}

fn main() {
    run();
}