**Description**

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 have to find the number of such integers * x * (* l * ≤ * x * ≤ * R *), that the first digit of Integer * x * & Nbsp;equals the last one (in decimal notation). For example, such numbers as 101, 477474 or < Span class= "Tex-span" >9 will be included in the answer And47, 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* ≤10 18)-the boundaries of the interval.

%lld specifier to read or write 64-bit integers inс++. 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.

**Examples**
**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, one,,.

Test instructions: With this number, the first and the last one are the same, asking you how many numbers meet the required number in the two number range ~

Solution: Of course, is to find the law, it is easy to see the law is 9*10^n, and then consider the situation of two endpoints, they are not also meet the requirements (here the code is not clear Qaq, the most valuable only the CMD function)

#include <bits/stdc++.h>using namespace Std;long long a[100005];int pos;int n;int d;long long cmd (string s) {long Long sum1=0,sum2=0; if (S.length () ==1) {sum1=0; } else if (S.length () ==2) {sum1=9; } else {sum1=10; for (int i=1;i<s.length () -2;i++) {sum1*=10; }//cout<<sum1<<endl; Sum1=sum1-1; cout<<sum1<<endl; sum1+=9; } int ans1=s[0]-' 0 '-1; Long Long sum3=0; Long Long pos1=1; if (S.length () >=3) {for (int i=1; I<=s.length ()-2; i++) {sum3=sum3*10+ (s[i]-' 0 '); }//Sum3=sum3+1; for (int i=0; I<s.length ()-2; i++) {pos1*=10; } if ((s[0]-' 0 ') <= (s[s.length () -1]-' 0 ')) {sum1=sum1+ans1*pos1+sum3+1; } else {sum1=sum1+ans1*pos1+sum3; }} else {if (S[0]<=s[s.length ()-1]){sum1=sum1+ans1*pos1+sum3+1; } else {sum1=sum1+ans1*pos1+sum3; }} return sum1;} int main () {string s1,s2; cin>>s1>>s2; Cout<<cmd (S1) <<endl; Cout<<cmd (S2) <<endl; if (S1[0]==s1[s1.length () -1]&&s2[0]<=s2[s2.length ()-1]) {cout<<cmd (s2)-cmd (S1) +1<<endl; } else if (S1[0]==s1[s1.length () -1]&&s2[0]>s2[s2.length ()-1]) {cout<<cmd (s2)-cmd (S1) +1< <endl; } else if (S1[0]<=s1[s1.length () -1]&&s2[0]<=s2[s2.length ()-1]) {cout<<cmd (s2)-cmd (S1) < <endl; } else if (S1[0]<=s1[s1.length () -1]&&s2[0]>s2[s2.length ()-1]) {cout<<cmd (s2)-cmd (S1) <& Lt;endl; } else if (S1[0]>s1[s1.length () -1]&&s2[0]<=s2[s2.length ()-1]) {cout<<cmd (s2)-cmd (S1) <& Lt;endl; } else if (S1[0]>s1[s1.length () -1]&&s2[0]>s2[s2.leNgth ()-1]) {cout<<cmd (s2)-cmd (S1) <<endl; } else if (S1[0]>s1[s1.length () -1]&&s2[0]<=s2[s2.length ()-1]) {cout<<cmd (s2)-cmd (S1) <& Lt;endl; } return 0;}

Codeforces Round #129 (Div. 2) C