Problem:
Score : 100 points

Problem Statement

Find the number of integers between 1 and K (inclusive) satisfying the following condition, modulo 10^9 + 7:

The sum of the digits in base ten is a multiple of D.

Constraints

All values in input are integers.

1 \leq K < 10^{10000}

1 \leq D \leq 100

Input

Input is given from Standard Input in the following format:

K
D

Output

Print the number of integers satisfying the condition, modulo 10^9 + 7.

Sample Input 1

30
4

Sample Output 1

6

Those six integers are: 4, 8, 13, 17, 22 and 26.

Sample Input 2

1000000009
1

Sample Output 2

2

Be sure to print the number modulo 10^9 + 7.

Sample Input 3

98765432109876543210
58

Sample Output 3

635270834

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#include <algorithm>
#include <cassert>
#include <cfloat>
#include <complex>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <regex>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <vector>

using namespace std;

#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define rep(i, n) for (ll i = 0; i < (n); ++i)
#define repLRE(i, l, r) for (ll i = (l); i <= (r); ++i)
#define rrepLRE(i, l, r) for (ll i = (l); i >= (r); --i)
#define Sort(v) sort(v.begin(), v.end())
#define rSort(v) sort(v.rbegin(), v.rend())
#define Reverse(v) reverse(v.begin(), v.end())
#define Lower_bound(v, x) \
  distance(v.begin(), lower_bound(v.begin(), v.end(), x))
#define Upper_bound(v, x) \
  distance(v.begin(), upper_bound(v.begin(), v.end(), x))

using ll = long long;
using ull = unsigned long long;
using P = pair<ll, ll>;
using T = tuple<ll, ll, ll>;
using vll = vector<ll>;
using vP = vector<P>;
using vT = vector<T>;
using vvll = vector<vector<ll>>;
using vvP = vector<vector<P>>;
using dqll = deque<ll>;

ll dx[9] = {-1, 1, 0, 0, -1, -1, 1, 1, 0};
ll dy[9] = {0, 0, -1, 1, -1, 1, -1, 1, 0};

/* Macros reg. ends here */

const ll INF = 1LL << 50;

static const long long mod = 1000000007;

int main() {
  // ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);

  string k;
  ll d;
  cin >> k >> d;

  ll n = k.size();

  vector<vvll> dp(2, vvll(n + 1, vll(d, 0)));
  dp[0][0][0] = 1;
  dp[1][0][0] = 0;
  repLRE(i, 1, n) {
    rep(j, d) {
      ll kid = k[i-1] - '0';
      dp[0][i][(j + kid) % d] = dp[0][i-1][j];
      rep(l, 10) {
        ll nj = (j + l) % d;
        dp[1][i][nj] = (dp[1][i][nj] + dp[1][i-1][j]) % mod;
      }  // d
      if (kid == 0) continue;
      repLRE(l, 0, kid - 1) {
        ll nj = (j + l) % d;
        dp[1][i][nj] = (dp[1][i][nj] + dp[0][i-1][j]) % mod;
      }  // d
    }    // j
  }      // i

  cout << dp[0][n][0] + dp[1][n][0] - 1 << endl;

  return 0;
}