#include <iostream>
#include <vector>
#include <string>
using namespace std;
int solve(string s) {
    if (s.empty() || s[0] == '0') {
        return 0;  
    }
    
    int n = s.size();
    vector<int> dp(n + 1, 0); 
    dp[0] = 1;  
    dp[1] = 1; 
    
    for (int i = 2; i <= n; ++i) {
        if (s[i - 1] >= '1' && s[i - 1] <= '9') {
            dp[i] += dp[i - 1];
        }
        int t = stoi(s.substr(i - 2, 2));
        if (t >= 10 && t <= 26) {
            dp[i] += dp[i - 2];
        }
    }
    
    return dp[n]; 
}

int main() {
    string s;
    cin >> s;
    int result = solve(s);
    cout << result << endl;

    return 0;
}

import re
a = ["Inc.", "Corp.", "LLC", "L.L.C.", "LLC."]
b = ["the", "an", "a", "and"]
def normalize_name(c):
    c = c.lower()
    for d in a:
        if c.endswith(d.lower()):
            c = c[:-len(d)].strip()
    c = re.sub(r'[&,]', ' ', c) 
    c = re.sub(r'\s+', ' ', c)  
    d = c.split()
    if d[0] in b:
        d = d[1:]
    c = ' '.join(d)
    return c.strip()
def check_availability(d):
    e = set()  
    for f in d:
        g, c = f.split('|')
        h = normalize_name(c)
        if not h:
            print(f"{g}|Name Not Available")
        elif h in e:
            print(f"{g}|Name Not Available")
        else:
            e.add(h)
            print(f"{g}|Name Available")

def process_string(number):
    while True:
        found_consecutive = False
        new_string = []
        i = 0
        while i < len(number):
            j = i
            while j < len(number) and number[j] == number[i]:
                j += 1
            if j - i > 1:
                found_consecutive = True
                sum_of_digits = sum(int(number[k]) for k in range(i, j))
                new_string.append(str(sum_of_digits))
            else:
                new_string.append(number[i])
            i = j
        number = ''.join(new_string)
        if not found_consecutive:
            break
    return number