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>
const int p = 1000000007;
struct matrix {
  int g[30][30];
  int s[30];
};
matrix G[80][30], H1[30], H2[30], S, T;
int h[80];
long long n;
int k, m, s;
inline void buildfirstmatrix() {
  scanf("%I64d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < m; j++) G[0][i].g[j][j] = 1;
    for (int j = 0; j < m; j++) G[0][i].g[i][j] = 1;
    G[0][i].s[i] = 1;
  }
  return;
}
inline void matrixmul(matrix& x, matrix& y) {
  memset(T.g, 0, sizeof(T.g));
  for (int i = 0; i < m; i++) T.s[i] = x.s[i];
  for (int i = 0; i < m; i++)
    for (int k = 0; k < m; k++)
      if (x.g[i][k]) {
        for (int j = 0; j < m; j++)
          T.g[i][j] = (T.g[i][j] + 1LL * x.g[i][k] * y.g[k][j]) % p;
        T.s[i] = (T.s[i] + 1LL * x.g[i][k] * y.s[k]) % p;
      }
  return;
}
inline void matrixpower(long long n) {
  h[k] = n % m, n = n / m;
  while (n > 0) {
    k++, h[k] = n % m, n = n / m;
    for (int i = 0; i < m; i++) H1[i] = H2[i] = G[k - 1][i];
    for (int i = 1; i < m; i++) matrixmul(H1[i - 1], H1[i]), H1[i] = T;
    for (int i = m - 2; i >= 0; i--) matrixmul(H2[i], H2[i + 1]), H2[i] = T;
    G[k][0] = H2[0];
    for (int i = 1; i < m; i++) matrixmul(H2[i], H1[i - 1]), G[k][i] = T;
  }
  for (int i = 0; i < m; i++) S.g[i][i] = 1;
  for (int i = k, l = 0; i >= 0; i--)
    for (int j = 0; j < h[i]; j++, l = (l + 1) % m)
      matrixmul(S, G[i][l]), S = T;
  for (int i = 0; i < m; i++) s = (s + S.s[i]) % p;
  printf("%d", (s + 1) % p);
  return;
}
int main() {
  buildfirstmatrix();
  matrixpower(n);
  return 0;
}