**Little Elephant and Interval

 Little Elephant and Interval

The Little Elephant very much loves sums on intervals.

This time he has a pair of integers l and r (l ≤ r). The Little Elephant has to find the number of such integers x (l ≤ x ≤ r), that the first digit of integer x equals the last one (in decimal notation). For example, such numbers as 101, 477474 or 9 will be included in the answer and 47, 253 or 1020 will not.

Help him and count the number of described numbers x for a given pair l and r.

Input

The single line contains a pair of integers l and r (1 ≤ l ≤ r ≤ 1018) — the boundaries of the interval.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use cin, cout streams or the %I64dspecifier.

Output

On a single line print a single integer — the answer to the problem.

Sample test(s)

input

2 47

output

12

input

47 1024

output

98

Note

In the first sample the answer includes integers 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44.

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this type of questions can be done with the technique f(l,r)=f(1,r)-f(1,l-1) Now suppose we have the number r=85546 Here we can go up to combinations like 7_ _ _ 7 with all possible permutations. But when we encounter the digit at MSB , we see the combination is formed like 8 _ _ _ 8 so there are 554 ways to normally fill this 3 dashed but as LSB<MSB so the only combonation that cannot be done is 85548 so we subtract 1 . Had the number been 8 _ _ _ 9 we could have added 554 combinations in between .

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#include<bits/stdc++.h> typedef long long int LL; using namespace std; LL calc(LL a) { if(a<10) return a; int ld=a%10; LL res=9+(a-10)/10;// 9 is added since numbers from 1-9 is consider special; while(a>9) a/=10; if(ld>=a) // if lsb>=msb

++res; return res; } int main() { //clock_t start=clock(); LL a,b; cin >> a >> b; cout << calc(b)-calc(a-1) <<endl; //fprintf(stderr,"time=%.3lfsec\n",0.000001*(clock()-start)); return 0; }

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It is well-known that for such problem you need to write function F(x) which solves the problem for the interval 0..x, and the answer then is F(r) - F(l - 1).

Now you need to write F(x) function. If x < 10, then answer is, of course, equal to x. Otherwise, let len be the length of x, x' — the integer x but without first and last digits, xi — the i-th digit of integer x (from left to right, starting from 0). Interate through all possible first digit d (which is the last at the same time) and through the length i of the number. Then if i < len - 2 or (i = len - 2 and d < x0) you need to add 10i to the answer. Otherwise, if i = len - 2 and d = x0 you need to add x' to the answer. Finally, if i = len - 2 and d = x0and xlen - 1 ≥ d add 1 to the answer.