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;
bool debug = 1;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
struct Matrix {
  int n = 32, m = 32;
  long long a[40][40];
  Matrix() {}
  Matrix(int n, int m) : n(n), m(m) {}
  void init(long long num) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) a[i][j] = num;
  }
  Matrix operator*(Matrix b) {
    Matrix o = Matrix(n, b.m);
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < b.m; j++) {
        o.a[i][j] = 0;
        for (int k = 0; k < m; k++) {
          o.a[i][j] =
              (o.a[i][j] + (a[i][k] * b.a[k][j]) % 1000000007) % 1000000007;
        }
      }
    }
    return o;
  }
};
int len, num[100];
Matrix f[70][35], pre[35], suf[35], ans;
int main() {
  scanf("%lld%d", &ln, &k);
  for (; ln; ln /= k) num[len++] = ln % k;
  ans.a[0][k] = 1;
  for (int(i) = 0; (i) < (int)(k); (i)++)
    for (int j = 0; j <= k; j++) {
      f[0][i].a[j][i] = 1;
      f[0][i].a[j][j] = 1;
    }
  for (int i = 1; i < len; i++) {
    pre[0] = f[i - 1][0];
    for (int(j) = 1; (j) <= (int)(k - 1); (j)++)
      pre[j] = pre[j - 1] * f[i - 1][j];
    suf[k - 1] = f[i - 1][k - 1];
    for (int j = k - 2; j >= 0; j--) {
      suf[j] = f[i - 1][j] * suf[j + 1];
    }
    f[i][0] = suf[0];
    for (int j = 1; j < k; j++) {
      f[i][j] = suf[j] * pre[j - 1];
    }
  }
  for (int i = len, s = 0; i >= 0; s += num[i], i--) {
    for (int j = 0; j < num[i]; j++) {
      ans = ans * f[i][(j + s) % k];
    }
  }
  int rs = 0;
  for (int i = 0; i <= k; i++) {
    (rs += ans.a[0][i]) %= 1000000007;
  }
  printf("%d\n", rs);
  return 0;
}