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;
inline void read(int &x) {
  int v = 0, f = 1;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void read(long long &x) {
  long long v = 0ll, f = 1ll;
  char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-')
    f = -1;
  else
    v = (c & 15);
  while (isdigit(c = getchar())) v = (v << 1) + (v << 3) + (c & 15);
  x = v * f;
}
inline void readc(char &x) {
  char c;
  while (((c = getchar()) == ' ') || c == '\n')
    ;
  x = c;
}
const int mod = 1e9 + 7;
long long n, m, i, j, dig[75], cnt;
struct mat {
  long long a[35][35];
} f[75][35], g, pre[35], suf[35];
mat mul(mat x, mat y) {
  mat z;
  int i, j, k;
  for ((i) = (0); (i) <= (m); (i)++) {
    for ((j) = (0); (j) <= (m); (j)++) {
      z.a[i][j] = 0;
      for ((k) = (0); (k) <= (m); (k)++) {
        z.a[i][j] = (z.a[i][j] + 1ll * x.a[i][k] * y.a[k][j]) % mod;
      }
    }
  }
  return z;
}
mat pw(mat x, long long y) {
  mat z;
  int i, j;
  for ((i) = (0); (i) <= (m); (i)++)
    for ((j) = (0); (j) <= (m); (j)++) {
      z.a[i][j] = (i == j);
    }
  while (y) {
    if (y & 1) {
      z = mul(z, x);
    }
    x = mul(x, x);
    y /= 2;
  }
  return z;
}
int main() {
  read(n);
  read(m);
  long long tmp = n;
  cnt = -1;
  while (tmp) {
    dig[++cnt] = tmp % m;
    tmp /= m;
  }
  g.a[0][m] = 1;
  for (((i)) = (0); ((i)) <= ((m)-1); ((i))++)
    for (((j)) = (0); ((j)) <= ((m + 1) - 1); ((j))++)
      f[0][i].a[j][i] = f[0][i].a[j][j] = 1;
  for (((i)) = (1); ((i)) <= ((cnt)); ((i))++) {
    pre[0] = f[i - 1][0];
    for (((j)) = (1); ((j)) <= ((m - 1)); ((j))++) {
      pre[j] = mul(pre[j - 1], f[i - 1][j]);
    }
    suf[m - 1] = f[i - 1][m - 1];
    for ((j) = (m - 2); (j) >= (0); (j)--) {
      suf[j] = mul(f[i - 1][j], suf[j + 1]);
    }
    f[i][0] = suf[0];
    for (((j)) = (1); ((j)) <= ((m - 1)); ((j))++)
      f[i][j] = mul(suf[j], pre[j - 1]);
  }
  int cur = 0;
  for ((i) = (cnt); (i) >= (0); (i)--) {
    while (dig[i]--) {
      g = mul(g, f[i][cur]);
      cur = (cur + 1) % m;
    }
  }
  int ans = 0;
  for (((i)) = (0); ((i)) <= ((m + 1) - 1); ((i))++)
    ans = (ans + g.a[0][i]) % mod;
  cout << ans << endl;
  return 0;
}