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;
int n, m, a[100];
struct mat {
  int a[35][35];
  mat() { memset(a, 0, sizeof(a)); }
  friend mat operator*(mat a, mat b) {
    mat c;
    for (int i = 0; i <= m; i++)
      for (int j = 0; j <= m; j++)
        for (int k = 0; k <= m; k++)
          c.a[i][j] = (c.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % 1000000007) %
                      1000000007;
    return c;
  }
} f, g, tmp, t, I, A;
void solve(int now) {
  if (now == 1) {
    for (int i = 0; i < m; i++) A.a[i][i] = 1;
    tmp = I;
    for (int i = 0; i < m; i++) {
      t = I;
      for (int j = 0; j <= m; j++)
        if (j != i) t.a[i][j]++;
      tmp = tmp * t;
      if (i == a[now]) f = tmp;
    }
    g = tmp;
  } else {
    solve(now - 1);
    tmp = I;
    for (int i = 0; i < a[now]; i++) {
      for (int j = 0; j <= m; j++)
        for (int k = 0; k <= m; k++) t.a[j][k] = 0;
      for (int j = 0; j < m; j++)
        for (int k = 0; k < m; k++) t.a[(j + i) % m][(k + i) % m] = g.a[j][k];
      for (int j = 0; j < m; j++) t.a[(j + i) % m][m] = g.a[j][m];
      t.a[m][m] = g.a[m][m];
      tmp = tmp * t;
    }
    for (int j = 0; j <= m; j++)
      for (int k = 0; k <= m; k++) t.a[j][k] = 0;
    for (int j = 0; j < m; j++)
      for (int k = 0; k < m; k++)
        t.a[(j + a[now]) % m][(k + a[now]) % m] = f.a[j][k];
    for (int j = 0; j < m; j++) t.a[(j + a[now]) % m][m] = f.a[j][m];
    t.a[m][m] = g.a[m][m];
    tmp = tmp * t;
    f = tmp;
    tmp = I;
    for (int i = 0; i < m; i++) {
      for (int j = 0; j <= m; j++)
        for (int k = 0; k <= m; k++) t.a[j][k] = 0;
      for (int j = 0; j < m; j++)
        for (int k = 0; k < m; k++) t.a[(j + i) % m][(k + i) % m] = g.a[j][k];
      for (int j = 0; j < m; j++) t.a[(j + i) % m][m] = g.a[j][m];
      t.a[m][m] = g.a[m][m];
      tmp = tmp * t;
    }
    g = tmp;
  }
}
int main() {
  long long t;
  scanf("%lld%d", &t, &m);
  t--;
  if (t == 0) {
    printf("2\n");
    return 0;
  }
  while (t) {
    a[++n] = t % m;
    t /= m;
  }
  for (int i = 0; i <= m; i++) I.a[i][i] = 1;
  solve(n);
  int ans = 1;
  for (int i = 0; i < m; i++) ans = (ans + f.a[i][m]) % 1000000007;
  printf("%d\n", ans);
  return 0;
}