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;
long long t1[50][50];
long long t[50][50];
long long tr2[50][50];
long long tmp[50][50];
long long x[50][50];
long long pmod = 1000000007;
int i, j, m;
void mul(long long a[][50], long long b[][50]) {
  int i, j, k;
  for (i = 0; i < m; ++i) {
    for (j = 0; j < m; ++j) {
      tmp[i][j] = 0;
      for (k = 0; k < m; ++k) {
        tmp[i][j] += a[k][j] * b[i][k] % pmod;
        tmp[i][j] %= pmod;
      }
    }
  }
  for (i = 0; i < m; ++i) {
    for (j = 0; j < m; ++j) {
      a[i][j] = tmp[i][j];
    }
  }
}
int main() {
  long long n;
  int i, j, k;
  scanf("%I64d%d", &n, &k);
  m = k + 1;
  memset(t1, 0, sizeof(t1));
  memset(tr2, 0, sizeof(tr2));
  memset(t, 0, sizeof(t));
  for (i = 0; i < k; ++i) {
    t1[i][(i + 1 + k) % k] = 1;
  }
  t1[i][i] = 1;
  for (i = 0; i <= k; ++i) {
    t[i][i] = 1;
  }
  for (i = 0; i <= k; ++i) {
    tr2[i][i] = 1;
    tr2[i][0] = 1;
  }
  while (n) {
    while (n % k != 0) {
      mul(t, t1);
      mul(t, tr2);
      n--;
    }
    n /= k;
    int kk = k;
    memset(x, 0, sizeof(x));
    for (i = 0; i <= k; ++i) {
      x[i][i] = 1;
    }
    while (kk--) {
      mul(x, t1);
      mul(x, tr2);
    }
    for (i = 0; i < m; ++i) {
      for (j = 0; j < m; ++j) {
        tr2[i][j] = x[i][j];
      }
    }
  }
  long long ans = 1;
  for (i = 0; i < k; ++i) {
    ans += t[k][i];
    ans %= pmod;
  }
  printf("%I64d\n", ans);
}