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++ (GCC 9.2.1) (judge verdict: Accepted)

#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>; 
using pll = pair<long long, long long>;
constexpr char ln =  '\n';
constexpr long long MOD = 1000000007LL;
constexpr long long INF = 1001001001LL;
constexpr long long LINF = 1001001001001001001;
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int i=0;i<(n);i++)
#define rept(i, j, n) for(int i=(j); i<(n); i++)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }


int main(){
    int a, b; cin >> a >> b;
    int cnt = 0;
    rept(i, a, b+1){
        string s = to_string(i);
        string t = s; reverse(all(t));
        if(s==t)cnt++;
    }
    cout << cnt << ln;
}