its>
#include <sstream>
using namespace std;

/*
brute force:
choice of range: O(n^2)
check each range if interesting: O(n)
total: O(n^3)

A better way:
scan from left to right, suppose we are at index i
total:
  total number of interesting range between [0, i-1]
prev_even:
  total number of ranges ending at i - 1 that contains even number of palindromes
prev_odd:
  total number of ranges ending at i - 1 that contains odd nubmer of palindromes

so
  If number at i is palindrome
    cur_even = prev_odd
    cur_odd = prev_even + 1
  else
    cur_even = prev_even + 1
    cur_odd = prev_odd
  total = toal + cur_even

As we spend O(x) time at each index, so total time O(nx), space O(1).
x is the time to check if a number is palindrome.
*/

bool is_palindrome(int x) {
    if (x <= 0) return false;

    long long y = x, z = 0;

    while (y > 0) {
        z = 10 * z + y % 10;
        y /= 10;
    }

    return z <= INT_MAX && z == x;
}

int solve(int begin, int end) {
    int total = 0;
    int prev_odd = 0, prev_even = 0, cur_odd = 0, cur_even = 0;

    for (int n = begin; n <= end; ++n) {
        if (is_palindrome(n)) {
            cur_even = prev_odd;
            cur_odd = prev_even + 1;
        } else {
            cur_even = prev_even + 1;
            cur_odd = prev_odd;
        }
        prev_even = cur_even;
        prev_odd = cur_odd;
        total += cur_even;
    }

    return total;
}

int main() {
    string line;
    int begin = 0, end = 0;

    while (getline(cin, line)) {
        istringstream iss(line);
        iss >> begin;
        iss >> end;
        cout << solve(begin, end) << endl;
    }

    return 0;
}