Problem: Assume that sk(n) equals the sum of digits of number n in the k-based notation. For example, s2(5) = s2(1012) = 1 + 0 + 1 = 2, s3(14) = s3(1123) = 1 + 1 + 2 = 4.

The sequence of integers a0, ..., an - 1 is defined as <image>. Your task is to calculate the number of distinct subsequences of sequence a0, ..., an - 1. Calculate the answer modulo 109 + 7.

Sequence a1, ..., ak is called to be a subsequence of sequence b1, ..., bl, if there is a sequence of indices 1 ≤ i1 < ... < ik ≤ l, such that a1 = bi1, ..., ak = bik. In particular, an empty sequence (i.e. the sequence consisting of zero elements) is a subsequence of any sequence.

Input

The first line contains two space-separated numbers n and k (1 ≤ n ≤ 1018, 2 ≤ k ≤ 30).

Output

In a single line print the answer to the problem modulo 109 + 7.

Examples

Input

4 2


Output

11


Input

7 7


Output

128

Note

In the first sample the sequence ai looks as follows: (0, 1, 1, 0). All the possible subsequences are: 

(), (0), (0, 0), (0, 1), (0, 1, 0), (0, 1, 1), (0, 1, 1, 0), (1), (1, 0), (1, 1), (1, 1, 0).

In the second sample the sequence ai looks as follows: (0, 1, 2, 3, 4, 5, 6). The subsequences of this sequence are exactly all increasing sequences formed from numbers from 0 to 6. It is easy to see that there are 27 = 128 such sequences.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename C>
void ma(C& a, C b) {
  if (a < b) a = b;
}
template <typename C>
void mi(C& a, C b) {
  if (a > b) a = b;
}
const int P = 1e9 + 7;
int n;
long long pom[31][31];
struct m {
  long long t[31][31];
  void id() {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) t[i][j] = (i == j);
  }
  void operator*=(m& b) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) pom[i][j] = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        for (int k = 0; k < n; k++) pom[i][k] += t[i][j] * b.t[j][k] % P;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) t[i][j] = pom[i][j] % P;
  }
  m(){};
  m(m b, int w) {
    id();
    while (w) {
      if (w & 1) *this *= b;
      w /= 2;
      b *= b;
    }
  }
} obr, nl, ak;
vector<m> d;
long long mm;
long long pw[31];
long long wyn[31];
int k;
vector<int> roz;
int main() {
  scanf("%lld%d", &mm, &k);
  while (mm) {
    roz.push_back(mm % k);
    mm /= k;
  }
  n = k + 1;
  for (int i = 0; i < k; i++) obr.t[i][(i + 1) % k] = 1;
  obr.t[k][k] = 1;
  for (int i = 0; i < k + 1; i++) {
    nl.t[0][i] = 1;
    nl.t[i][i] = 1;
  }
  d.push_back(nl);
  d[0] *= obr;
  for (int i = 1; i < roz.size(); i++) {
    d.push_back(m(d[i - 1], k));
    d[i] *= obr;
  }
  for (int i = 0; i < n; i++) wyn[i] = 1;
  for (int ii = roz.size() - 1; ii >= 0; ii--) {
    m pom(d[ii], roz[ii]);
    for (int i = 0; i < n; i++) pw[i] = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) pw[j] += wyn[i] * pom.t[i][j] % P;
    for (int i = 0; i < n; i++) wyn[i] = pw[i] % P;
  }
  printf("%lld\n", wyn[k]);
}