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 <bits/stdc++.h>

using namespace std;

const double PI = 3.141592653589;
#define fastio ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pb push_back
#define mp make_pair
#define int long long
#define ll long long
#define ll long long
#define all(c) (c).begin(),(c).end()
#define M 1000000007
#define INF LLONG_MAX
#define pr(...) dbs(#__VA_ARGS__, __VA_ARGS__)
template <class T> void dbs(string str, T t) {cerr << str << " : " << t << "\n";}
template <class T, class... S> void dbs(string str, T t, S... s) {int idx = str.find(','); cerr << str.substr(0, idx) << " : " << t << ", "; dbs(str.substr(idx + 1), s...);}
template <class S, class T>ostream& operator <<(ostream& os, const pair<S, T>& p) {return os << "(" << p.first << ", " << p.second << ")";}
template <class T>ostream& operator <<(ostream& os, const vector<T>& p) {os << "[ "; for (auto& it : p) os << it << " "; return os << "]";}
template <class T>ostream& operator <<(ostream& os, const set<T>& p) {os << "[ "; for (auto& it : p) os << it << " "; return os << "]";}
template <class S, class T>ostream& operator <<(ostream& os, const map<S, T>& p) {os << "[ "; for (auto& it : p) os << it << " "; return os << "]";}
template <class T> void prc(T a, T b) {cerr << "["; for (T i = a; i != b; ++i) {if (i != a) cerr << ", "; cerr << *i;} cerr << "]\n";}
// Use pr(a,b,c,d,e) or cerr<<anything or prc(v.begin(),v.end()) or prc(v,v+n)
//  


int d;
string k;
int dp[10001][101][2]; // index sum%d tight
const int MOD = 1e9 + 7;
int f(int index, int sum, bool tight)
{
    sum %= d;
    if(index == (int)k.size()) 
    {
        if(sum%d == 0) return 1;
        return 0;
    }
    if(dp[index][sum][tight] != -1) return dp[index][sum][tight];
    int last = 9;
    if(tight) last = k[index] - '0';
    int curr =0;
    for(int i=0;i<=last;i++)
    {
        if(tight and i == last)
            curr += f(index+1, sum+i, tight);
        else
            curr += f(index+1, sum+i, 0);
    }

    dp[index][sum][tight] = curr%MOD;
    return curr%MOD;
}

int32_t main()
{
    fastio;
    //freopen("file.in", "r", stdin);
    //freopen("file.out", "w", stdout);
    cin >> k >> d;
    memset(dp, -1, sizeof(dp));
    cout << f(0, 0, 1) - 1 << "\n"; // -1 for 0
    return 0;
}