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)

// #pragma GCC optimize("O3")
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vll = vector<ll>;
using vii = vector<int>;
using pii = pair<int, int>;

const ll INF = numeric_limits<ll>::max(),
         MOD = 1e9 + 7;

const int INF_i = numeric_limits<int>::max(),
          MAX = 1e6;

int main() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);

    string s;
    cin >> s;
    int D;
    cin >> D;
    vll dp(D);

    int prev = s[0] - '0';
    for (int d = 1; d < prev; d++) {
        dp[d % D] = 1;
    }

    for (int i = 1; i < s.size(); i++) {
        int di = s[i] - '0';
        vll prev_dp = dp;
        dp.assign(D, 0);
        for (int d = 0; d < 10; d++) {
            if (d) (++dp[d % D]) %= MOD;
            for (int j = 0; j < D; j++) {
                (dp[(j + d) % D] += prev_dp[j]) %= MOD;
            }
            if (d < di) (++dp[(prev + d) % D]) %= MOD;
        }

        (prev += di) %= D;
    }
    cout << (dp[0] + (!prev)) % MOD << '\n';
}