#include<bits/stdc++.h>
using namespace std;
const int MOD = 1e9+7;
char a[10010];
int dp[10010][2][110];
int n,len;
int play(int now,int state,int num){
    if(dp[now][state][num]!=-1) return dp[now][state][num];
    if(now == len)    return (num == 0);
    int sum = 0;
    int ma = (state)?a[now]-'0':9;
    for(int i=0;i<=ma;i++)
        sum = (sum + play(now+1,state && (i==ma),(num+i)%n ) )%MOD;
    return dp[now][state][num] = sum;
}
int main(){
    scanf(" %s %d",a,&n);
    len = strlen(a);
    memset(dp,-1,sizeof dp);
    printf("%d\n",(play(0,1,0)-1+MOD)%MOD);
    return 0;
}

### Prompt

Please create a solution in CPP to 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;
const int MOD = 1e9+7;
char a[10010];
int dp[10010][2][110];
int n,len;
int play(int now,int state,int num){
    if(dp[now][state][num]!=-1) return dp[now][state][num];
    if(now == len)    return (num == 0);
    int sum = 0;
    int ma = (state)?a[now]-'0':9;
    for(int i=0;i<=ma;i++)
        sum = (sum + play(now+1,state && (i==ma),(num+i)%n ) )%MOD;
    return dp[now][state][num] = sum;
}
int main(){
    scanf(" %s %d",a,&n);
    len = strlen(a);
    memset(dp,-1,sizeof dp);
    printf("%d\n",(play(0,1,0)-1+MOD)%MOD);
    return 0;
}

```