#include <bits/stdc++.h>
#define N 25
using namespace std;
typedef long long ll;
ll a, b, A[N], B[N], sz[2];
void fill(ll x, ll t)
{
vector<ll> v;
while(x > 0)
{
ll r = x % 10;
v.push_back(r);
x /= 10;
}
reverse(v.begin(), v.end());
sz[t] = v.size();
for(int i = 0; i < v.size(); i++)
if(!t) A[i+1] = v[i];
else B[i+1] = v[i];
}
// F = 1, O NUMERO JÁ É ESTRITAMENTE MENOR
//F = 0, O NUMERO É EXATAMENTE IGUAL
ll dp[N][13][13][2];
ll solveA(int i, int a, int b, int f)
{
if(i > sz[0]) return 1;
if(dp[i][a+1][b+1][f] != -1) return dp[i][a+1][b+1][f];
ll ans = 0;
for(int k = 0; k < 10; k++)
{
if((i == 1 and !k) or k == a or k == b) continue;
if(f == 1) ans += solveA(i + 1, k, a, f);
else
{
if(k < A[i]) ans += solveA(i + 1, k, a, 1);
else if(k == A[i]) ans += solveA(i + 1, k, a, 0);
}
}
return dp[i][a+1][b+1][f] = ans;
}
int t;
ll solveA2(int i, int a, int b, int f)
{
if(i > t) return 1;
if(dp[i][a+1][b+1][f] != -1) return dp[i][a+1][b+1][f];
ll ans = 0;
for(int k = 0; k < 10; k++)
{
if((i == 1 and !k) or k == a or k == b) continue;
ans += solveA2(i + 1, k, a, f);
}
return dp[i][a+1][b+1][f] = ans;
}
int main()
{
cin>>a>>b;
a--;
fill(a, 0);
memset(dp, -1, sizeof dp);
ll L = solveA(1, -1, -1, 0);
for(int T = 1; T < sz[0]; T++)
{
t = T;
memset(dp, -1, sizeof dp);
L += solveA2(1, -1, -1, 0);
}
memset(dp, -1, sizeof dp);
fill(b, 0);
ll R = solveA(1, -1, -1, 0);
for(int T = 1; T < sz[0]; T++)
{
t = T;
memset(dp, -1, sizeof dp);
R += solveA2(1, -1, -1, 0);
}
if(a < 0) L -= 2;
if(a == 0) L--;
if(b == 0) R --;
cout<<R - L<<"\n";
}
