#include <bits/stdc++.h>
using namespace std;
long long dp[20][10][10];
int v[20];
long long a,b;
int i,j,k,t;
long long solve (long long n){
if (n < 0)
return 0;
if (!n)
return 1;
long long nr = n, sol = 1;
int i,j,k,poz,el = 0;
memset (v,0,sizeof v);
while (nr){
v[++el] = nr%10;
nr /= 10;
}
reverse (v+1,v+el+1);
for (i=1;i<el;i++){
for (j=0;j<=9;j++)
for (k=1;k<=9;k++)
sol += dp[i][j][k];
}
for (poz=1,i=el;poz<=el;poz++,i--){
int start = (poz == 1) ? (1) : (0);
if (poz == el)
v[poz]++;
for (k=start;k<v[poz];k++){
if ( (poz > 1 && k == v[poz-1]) || (poz > 2 && k == v[poz-2]))
continue;
for (j=0;j<=9;j++)
if ((i == 1) || (j != v[poz-1] || poz == 1) )
sol += dp[i][j][k];
}
if ((poz > 1 && v[poz] == v[poz-1]) || (poz > 2 && v[poz] == v[poz-2]))
break;
}
return sol;
}
int main (){
//ifstream cin ("date.in");
//ofstream cout ("date.out");
/// dp[i][j][k] - in cate moduri pot sa construiesc un nr din i cifre cu ultimele doua j,k
for (i=0;i<=9;i++)
dp[1][0][i] = 1;
for (i=0;i<=9;i++)
for (j=0;j<=9;j++)
if (i != j)
dp[2][i][j] = 1;
for (i=3;i<=18;i++){
for (j=0;j<=9;j++){
for (k=0;k<=9;k++){
if (j == k)
continue;
for (t=0;t<=9;t++)
if (t != k && t != j)
dp[i][j][k] += dp[i-1][t][j];
}}}
cin>>a>>b;
cout<<solve (b) - solve (a-1);
return 0;
}
