#include <bits/stdc++.h>

using namespace std;
#define ll long long

const int MOD = 1e9+7;


string s; 
ll d, n;
ll dp[10005][101][2];  // k = 0, cannot cross limit, k = 1 can

ll solve(ll i, ll sum, ll k){
    if(i == n) return sum == 0;
    if(dp[i][sum][k] == -1) {
    ll dig = s[i] - '0';
    if(k) dig = 9;
    
ll res = 0;
    for(int j = 0; j <= dig; j++){
    res += solve(i+1, (sum + j)%d, (k == 1 ||  (k == 0 && j < dig ) ? 1 : 0)  );
    res %= MOD;
}
dp[i][sum][k] = res;
}
return dp[i][sum][k];
}
 
int main(){
    cin >> s >> d;
    n = s.size();
    
    memset(dp, -1, sizeof(dp));

    cout << (solve(0, 0, 0) - 1 + MOD)%MOD;
    
}



### Prompt

Create a solution in cpp for the following problem:

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.

Examples

Input

30
4


Output

6


Input

1000000009
1


Output

2


Input

98765432109876543210
58


Output

635270834

### Response

```cpp
#include <bits/stdc++.h>

using namespace std;
#define ll long long

const int MOD = 1e9+7;


string s; 
ll d, n;
ll dp[10005][101][2];  // k = 0, cannot cross limit, k = 1 can

ll solve(ll i, ll sum, ll k){
    if(i == n) return sum == 0;
    if(dp[i][sum][k] == -1) {
    ll dig = s[i] - '0';
    if(k) dig = 9;
    
ll res = 0;
    for(int j = 0; j <= dig; j++){
    res += solve(i+1, (sum + j)%d, (k == 1 ||  (k == 0 && j < dig ) ? 1 : 0)  );
    res %= MOD;
}
dp[i][sum][k] = res;
}
return dp[i][sum][k];
}
 
int main(){
    cin >> s >> d;
    n = s.size();
    
    memset(dp, -1, sizeof(dp));

    cout << (solve(0, 0, 0) - 1 + MOD)%MOD;
    
}



```