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#include <bits/stdc++.h>

#define FOR(i, b, e) for (int i = (b); i <= (e); i++)
#define FORD(i, b, e) for (int i = (e); i >= (b); i--)
#define MP make_pair
#define FS first
#define ND second
#define PB push_back
#define ALL(x) x.begin(), x.end()

using namespace std;

using LL = long long;
using PII = pair<int,int>;
using PLL = pair<LL,LL>;
using VI = vector<int>;

template<class T>
int sz(const T& a) { return (int)a.size(); }
template<class T>
void amin(T& a, T b) { a = min(a, b); }
template<class T>
void amax(T& a, T b) { a = max(a, b); }

const int inf = 1e9;
const LL infl = 1e18;

const int M = 5 * 7 * 8 * 9;
VI a;
LL dp[20][2][512][M];

LL solve(LL n)
{
  if (!n) return 0;
  a.clear();
  while (n) {
    a.PB(n % 10);
    n /= 10;
  }
  a.PB(0);
  reverse(ALL(a));

  memset(dp, 0, sizeof(dp));

  FOR(d, 1, a[1]) dp[1][d < a[1]][1 << (d - 1)][d] = 1;

  FOR(i, 1, sz(a) - 2) {
    FOR(f, 0, 1) FOR(d, 1, f ? 9 : a[i + 1]) FOR(m, 1, 511) FOR(x, 0, M - 1) {
      dp[i + 1][f | (d < a[i + 1])][m | (1 << (d - 1))][(10 * x + d) % M] += dp[i][f][m][x];
    }
    FOR(d, 1, 9) dp[i + 1][1][1 << (d - 1)][d]++;
  }

  LL ans = 0;
  FOR(f, 0, 1) FOR(m, 1, 511) FOR(x, 0, M - 1) {
    bool ok = true;
    FOR(d, 1, 9) ok &= (!(m & (1 << (d - 1))) || !(x % d));
    ans += ok * dp[sz(a) - 1][f][m][x];
  }
  return ans;
}

int main()
{
  LL l, h;
  scanf("%lld %lld", &l, &h);
  LL ans = solve(h) - solve(l - 1);
  printf("%lld\n", ans);
}