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 mod = 1e9 + 7;
using namespace std;
int m, lim[70], len = 0;
long long n;
struct mat {
  int n, m, a[31][31];
  mat(int _n = 0, int _m = 0) {
    n = _n, m = _m;
    memset(a, 0, sizeof(a));
  }
  int *operator[](int b) { return a[b]; }
  mat operator*(mat &b) const {
    assert(m == b.n);
    mat tmp(n, b.m);
    for (int i = 0; i <= n; i++)
      for (int k = 0; k <= m; k++)
        for (int j = 0; j <= b.m; j++)
          tmp[i][j] = (tmp[i][j] + 1ll * a[i][k] * b[k][j]) % mod;
    return tmp;
  }
  void print() {
    for (int i = 0; i <= n; i++, puts(""))
      for (int j = 0; j <= m; j++) printf("%d ", a[i][j]);
  }
} f[70][31], pre[70][31], suf[70][31];
void work(int dep, int nw, int sum, mat &res) {
  if (dep < 0) return;
  if (!nw) return work(dep - 1, dep ? lim[dep - 1] : 0, sum, res);
  work(dep, nw - 1, (sum + 1) % m, res);
  res = f[dep][sum] * res;
}
void qmo(int &x) { x += (x >> 31) & mod; }
int main() {
  scanf("%lld%d", &n, &m);
  while (n) lim[len++] = n % m, n /= m;
  for (int i = 0; i < len; i++) {
    for (int j = 0; j < m; j++) {
      f[i][j] = pre[i][j] = suf[i][j] = mat(m, m);
      if (!i)
        for (int k = 0; k <= m; k++) f[0][j][k][k] = 1, f[0][j][j][k] = 1;
      else if (!j)
        f[i][0] = suf[i - 1][0];
      else
        f[i][j] = suf[i - 1][j] * pre[i - 1][j - 1];
    }
    pre[i][0] = f[i][0], suf[i][m - 1] = f[i][m - 1];
    for (int j = 1; j < m; j++) pre[i][j] = pre[i][j - 1] * f[i][j];
    for (int j = m - 2; j >= 0; j--) suf[i][j] = f[i][j] * suf[i][j + 1];
  }
  mat res(m, 0);
  res[m][0] = 1;
  work(len - 1, lim[len - 1], 0, res);
  int ans = 0;
  for (int i = 0; i <= m; i++) qmo(ans += res[i][0] - mod);
  printf("%d\n", ans);
}