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;
inline void add(long long &x, long long y) { x = (x + y) % 1000000007ll; }
long long n;
int K, R;
struct M {
  long long a[35][35];
  void Z() { memset(a, 0, sizeof a); }
  void E() {
    Z();
    for (int i = 0; i < int(R); ++i) a[i][i] = 1;
  }
  friend M operator*(const M &x, const M &y) {
    M z;
    z.Z();
    for (int i = 0; i < int(R); ++i)
      for (int j = 0; j < int(R); ++j)
        for (int k = 0; k < int(R); ++k) add(z.a[i][j], x.a[i][k] * y.a[k][j]);
    return z;
  }
};
M r[35];
M o[70];
int main() {
  cin >> n >> K;
  R = K + 1;
  int N[70], B = 0;
  while (n) N[B++] = n % K, n /= K;
  r[0].E();
  r[1].Z();
  r[1].a[0][0] = 1;
  for (int j = 1; j <= int(K); ++j) r[1].a[j % K + 1][j] = 1;
  for (int i = 1; i <= int(K); ++i) r[i] = r[1] * r[i - 1];
  o[0].E();
  for (int i = 0; i < int(R); ++i) o[0].a[1][i] = 1;
  for (int i = 1; i <= int(B); ++i) {
    o[i].E();
    for (int j = 0; j < int(K); ++j) o[i] = r[j] * o[i - 1] * r[K - j] * o[i];
  }
  M res;
  res.E();
  int s = 0;
  for (int i = 0; i < int(B); ++i)
    for (int j = 0; j < int(N[B - 1 - i]); ++j)
      res = r[s % K] * o[B - 1 - i] * r[K - s % K] * res, s++;
  long long ans = 0;
  for (int i = 0; i < int(R); ++i) add(ans, res.a[i][0]);
  cout << ans << endl;
  return 0;
}