#include<cstdio>
#include<iostream>
#include<queue>
#include<algorithm>
#define ll long long
#define M 40009
#define MO 1000000007
using namespace std;
int a[],tot,K,anss;
ll ans[*M],n,f[][M][],size[M],m;
struct data
{
int x,y;
ll z;
data(int a1,int a2)
{
x=a1;
y=a2;
z=size[x]*size[y];
}
bool operator <(data a1)const
{
return z<a1.z;
}
};
void dfs(int wei,ll a1,int now)
{
if(a1>m)
return;
if(wei==tot)
{
ans[++ans[]]=a1;
return;
}
for(int i=now;i<;i++)
dfs(wei+,a1*i,i);
return;
}
bool cmp(ll a1,ll a2)
{
return a1>a2;
}
int main()
{
scanf("%lld%d",&n,&K);
m=n;
for(;n;n/=)
a[++tot]=n%;
dfs(,,);
sort(ans+,ans+ans[]+);
ans[]=unique(ans+,ans+ans[]+)-ans-;
f[][][]=;
for(int i=;i<tot;i++)
{
for(int j=;j<=ans[];j++)
for(int k=;k<;k++)
if(f[i][j][k])
for(int l=;l<;l++)
{
ll a1=ans[j]*l;
if(a1>ans[ans[]])
continue;
a1=lower_bound(ans+,ans+ans[]+,a1)-ans;
f[i+][a1][(k+l)>a[i+]]+=f[i][j][k];
}
}
for(int i=;i<ans[];i++)
for(int j=;j<tot;j++)
size[i]+=f[j][i][]+f[j][i][];
for(int i=;i<=ans[];i++)
size[i]+=f[tot][i][];
sort(size+,size+ans[]+,cmp);
priority_queue<data> p;
p.push(data(,));
for(;!p.empty();)
{
K--;
data q=p.top();
p.pop();
anss=(anss+q.z)%MO;
if(!K)
break;
if(q.x!=q.y)
{
anss=(anss+q.z)%MO;
K--;
if(!K)
break;
p.push(data(q.x+,q.y));
}
if(q.x==)
p.push(data(q.x,q.y+));
}
printf("%d",anss);
return ;
}
DP求出转移到x坐标的数目。各位数乘积的实际数目少。