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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
const int mod = 998244353;
typedef __int128 ll;
map <pair <ll, int>, int> ans;
inline int dfs(pair <ll, int> now) {
    if (now.second == 0) return 1;
    if (::ans.count(now)) return ::ans[now];
    int ans = dfs(make_pair(now.first >> 1, now.second - 1)) * ((now.first & 1) + 1) % mod;
    for (int k = 1; k <= now.second >> 1; k++) {
        ll S = ((ll)1 << k) - 1;
        ll tmp = now.first, sum = tmp & S;
        tmp >>= k;
        sum &= tmp & S;
        tmp >>= k;
        int rest = now.second - k - k;
        for (; rest >= 0; rest -= k, sum &= tmp & S, tmp >>= k) {
            ans = (ans + 1ll * dfs(make_pair(sum, k)) * dfs(make_pair(tmp, rest))) % mod;
        }
    }
    return ::ans[now] = ans;
}
char s[110];
int main() {
    scanf("%s", s);
    int n = strlen(s);
    __int128 S = 0;
    for (int i = 0; i < n; i++) S = S << 1 | (s[i] - '0');
    return cout << dfs(make_pair(S, n)) << endl, 0;
}