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;
const int MOD = 1e9 + 7;
long long n;
int m;
int tot;
int a[65];
inline int mom(int x) {
  if (x >= m) return x - m;
  return x;
}
inline int mo(int x) {
  if (x >= MOD) return x - MOD;
  return x;
}
struct mtx {
  int d[35][35];
  mtx() { memset(d, 0, sizeof(d)); }
  friend mtx operator*(const mtx &a, const mtx &b) {
    mtx c;
    for (int k = 0; k <= m; k++)
      for (int i = 0; i <= m; i++)
        for (int j = 0; j <= m; j++)
          c.d[i][j] = mo(c.d[i][j] + 1ll * a.d[i][k] * b.d[k][j] % MOD);
    return c;
  }
  friend mtx operator+(const mtx &a, const mtx &b) {
    mtx c;
    for (int i = 0; i <= m; i++)
      for (int j = 0; j <= m; j++) c.d[i][j] = mo(a.d[i][j] + b.d[i][j]);
    return c;
  }
  friend mtx operator-(const mtx &a, const mtx &b) {
    mtx c;
    for (int i = 0; i <= m; i++)
      for (int j = 0; j <= m; j++) c.d[i][j] = mo(a.d[i][j] - b.d[i][j] + MOD);
    return c;
  }
} ans, t[65][35], pre[35], suf[35];
int main() {
  scanf("%lld%d", &n, &m);
  ans.d[0][m] = 1;
  for (int i = 0; i < m; i++)
    for (int j = 0; j <= m; j++) t[0][i].d[j][i] = t[0][i].d[j][j] = 1;
  while (n) a[tot++] = n % m, n /= m;
  for (int i = 1; i < tot; i++) {
    pre[0] = t[i - 1][0];
    for (int j = 1; j < m; j++) pre[j] = pre[j - 1] * t[i - 1][j];
    suf[m - 1] = t[i - 1][m - 1];
    for (int j = m - 2; j >= 0; j--) suf[j] = t[i - 1][j] * suf[j + 1];
    t[i][0] = suf[0];
    for (int j = 1; j < m; j++) t[i][j] = suf[j] * pre[j - 1];
  }
  int sum = 0;
  for (int i = tot - 1; i >= 0; i--) {
    for (int j = 0; j < a[i]; j++) ans = ans * t[i][mom(sum + j)];
    sum = mom(sum + a[i]);
  }
  int anss = 0;
  for (int i = 0; i <= m; i++) anss = mo(anss + ans.d[0][i]);
  printf("%d\n", anss);
}