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;
struct M {
  unsigned long long int s[35][35], n;
  friend M operator*(M a, M b) {
    M c;
    c.n = a.n;
    int i, j, k;
    for (i = 0; i < a.n; i++) {
      for (j = 0; j < a.n; j++) {
        c.s[i][j] = 0;
        for (k = 0; k < a.n; k++) {
          c.s[i][j] += a.s[i][k] * b.s[k][j];
          c.s[i][j] %= 1000000007;
        }
      }
    }
    return c;
  }
} bi[65][65], I;
int main() {
  unsigned long long int n, m, t, x, y, z, add;
  int i, j, k;
  scanf("%llu%llu", &n, &m);
  bi[0][0].n = m + 1;
  for (i = 0; i <= m; i++) {
    bi[0][0].s[i][i] = 1;
    bi[0][0].s[0][i] = 1;
  }
  x = 0, y = 1, z = 0, add = 0;
  while (y * m <= n) {
    y *= m;
    x++;
  }
  for (t = 0; t <= x; t++) {
    if (t) {
      bi[t][0] = bi[t - 1][0];
      for (i = 1; i < m; i++) {
        bi[t][0] = bi[t][0] * bi[t - 1][i];
      }
    }
    for (i = 1; i < m; i++) {
      bi[t][i] = bi[t][i - 1];
      for (j = m - 2; j >= 0; j--) {
        for (k = 0; k <= m; k++) {
          swap(bi[t][i].s[j][k], bi[t][i].s[j + 1][k]);
        }
      }
      for (j = m - 2; j >= 0; j--) {
        for (k = 0; k <= m; k++) {
          swap(bi[t][i].s[k][j], bi[t][i].s[k][j + 1]);
        }
      }
    }
  }
  for (i = 0; i <= m; i++) {
    I.n = m + 1;
    I.s[i][i] = 1;
  }
  while (add < n) {
    if (add <= n - y) {
      add += y;
      I = I * bi[x][z];
      z = (z + 1) % m;
    } else {
      y /= m;
      x--;
    }
  }
  unsigned long long int ans = 0;
  for (i = 0; i <= m; i++) {
    ans = (ans + I.s[i][m]) % 1000000007;
  }
  printf("%llu\n", ans);
}