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;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
const int MOD = 1000000007;
const long long LIM = LLONG_MAX - (long long)MOD * MOD;
typedef int Mat[32][32];
long long n;
int base;
void mmult(const Mat &A, const Mat &B, Mat &C) {
  for (int i = (0); i < (base + 1); ++i)
    for (int j = (0); j < (base + 1); ++j) {
      long long cur = 0;
      for (int k = (0); k < (base + 1); ++k) {
        if (cur > LIM) cur %= MOD;
        cur += (long long)A[i][k] * B[k][j];
      }
      C[i][j] = cur % MOD;
    }
}
void mcopy(const Mat &A, Mat &B) {
  for (int i = (0); i < (base + 1); ++i)
    for (int j = (0); j < (base + 1); ++j) B[i][j] = A[i][j];
}
void printmat(const Mat &A) {
  for (int i = (0); i < (base + 1); ++i) {
    for (int j = (0); j < (base + 1); ++j) printf("%2d ", A[i][j]);
    puts("");
  }
}
Mat mat[60][32];
Mat have, tmp;
void run() {
  cin >> n >> base;
  for (int i = (0); i < (base); ++i)
    for (int j = (0); j < (base + 1); ++j)
      for (int k = (0); k < (base + 1); ++k)
        mat[0][i][j][k] = i == j || j == k ? 1 : 0;
  int ci = 0, cj = 0;
  long long step = 1;
  while (step <= n / base) ++ci, step *= base;
  for (int i = (1); i <= (ci); ++i)
    for (int j = (0); j < (base); ++j) {
      for (int k = (0); k < (base + 1); ++k)
        for (int l = (0); l < (base + 1); ++l) mat[i][j][k][l] = k == l ? 1 : 0;
      for (int k = (0); k < (base); ++k) {
        mmult(mat[i][j], mat[i - 1][(j + k) % base], tmp);
        mcopy(tmp, mat[i][j]);
      }
    }
  for (int k = (0); k < (base + 1); ++k)
    for (int l = (0); l < (base + 1); ++l) have[k][l] = k == l ? 1 : 0;
  for (; ci >= 0; --ci, step /= base) {
    int times = (n / step) % base;
    for (int x = (0); x < (times); ++x) {
      mmult(have, mat[ci][cj], tmp);
      mcopy(tmp, have);
      if (++cj >= base) cj -= base;
    }
  }
  int ret = 0;
  for (int k = (0); k < (base + 1); ++k) {
    ret += have[k][base];
    if (ret >= MOD) ret -= MOD;
  }
  printf("%d\n", ret);
}
int main() {
  run();
  return 0;
}