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 N = 31, S = 64;
int k;
const int mod = 1e9 + 7;
struct M {
  int a[N][N];
  inline M operator*(const M &b) const {
    M ans = {};
    for (int i = 0; i <= k; i++)
      for (int j = 0; j <= k; j++)
        for (int t = 0; t <= k; t++)
          ans.a[i][j] = (ans.a[i][j] + 1LL * a[i][t] * b.a[t][j]) % mod;
    return ans;
  }
} f[S][N], pre[N], suf[N];
int main() {
  long long n;
  int num[S], len = 0, i, j, s;
  M ans = {};
  cin >> n >> k;
  ans.a[0][k] = 1;
  while (n) num[len++] = n % k, n /= k;
  for (i = 0; i < k; i++)
    for (j = 0; j <= k; j++) f[0][i].a[j][j] = f[0][i].a[j][i] = 1;
  for (i = 1; i < len; i++) {
    pre[0] = f[i - 1][0];
    for (j = 1; j < k; j++) pre[j] = pre[j - 1] * f[i - 1][j];
    suf[k - 1] = f[i - 1][k - 1];
    for (j = k - 1; j--;) suf[j] = f[i - 1][j] * suf[j + 1];
    f[i][0] = suf[0];
    for (j = 1; j < k; j++) f[i][j] = suf[j] * pre[j - 1];
  }
  for (i = len, s = 0; i--; s += num[i])
    for (j = 0; j < num[i]; j++) ans = ans * f[i][(j + s) % k];
  s = 0;
  for (i = 0; i <= k; i++) (s += ans.a[0][i]) %= mod;
  printf("%d\n", s);
  return 0;
}