Problem:
Score : 200 points

Problem Statement

Find the number of palindromic numbers among the integers between A and B (inclusive).
Here, a palindromic number is a positive integer whose string representation in base 10 (without leading zeros) reads the same forward and backward.

Constraints

10000 \leq A \leq B \leq 99999

All input values are integers.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the number of palindromic numbers among the integers between A and B (inclusive).

Sample Input 1

11009 11332

Sample Output 1

4

There are four integers that satisfy the conditions: 11011, 11111, 11211 and 11311.

Sample Input 2

31415 92653

Sample Output 2

612

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

/*input
11009 11332
*/
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<int,int> pii; 


#define FileIn(file) freopen(file, "r", stdin)
#define FileOut(file) freopen(file, "w", stdout)
#define syso(n) cout<<n<<endl;
#define sysom(message, n) cout<<message<<n<<endl;
#define all(c) c.begin(), c.end() 
#define tr(container, it) \
for(typeof(container.begin()) it = container.begin(); it != container.end(); it++) 
#define present(container, element) (container.find(element) != container.end())  //map, set
#define cpresent(container, element) (find(all(container),element) != container.end()) //vector
#define what_is(x) cout << #x << " is " << x << endl; 


ll gcd(ll a, ll b){
    if (b==0) return a;
    return gcd(b, a%b);
} 

ll lcm(ll fir, ll sec){
    return (fir*sec)/(gcd(fir, sec));
}

//x^n%M
// log(n) time, log(n) memory
ll modularExponentiation(ll x,ll n,ll M)
{
    if(n==0)
        return 1;
    else if(n%2 == 0)        //n is even
        return modularExponentiation((x*x)%M,n/2,M);
    else                             //n is odd
        return (x*modularExponentiation((x*x)%M,(n-1)/2,M))%M;

}

const int MOD = 1e9 + 7;
const string small_alphabets = "abcdefghijklmnopqrstuvwxyz";
const int month_days[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
// --
// -- 
// -- 
// std::cout << std::fixed;
// std::cout << std::setprecision(5) << f << '\n';


// ios_base::sync_with_stdio(false);
// cin.tie(NULL);

bool pali(int x){
    // syso("here");
    string s = to_string(x);
    // what_is(s);
    string r = s;
    reverse(r.begin(), r.end());
    if(r==s){
        return true;
    }
    return false;
}

int main(){
    int a,b;
    cin>>a>>b;
    int ans = 0;
    for (int i = a; i <= b; ++i){
        if(pali(i)){
            ans++;
        }
    }
    syso(ans);
}