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>
const int P = 1000000007;
long long p = 1ll * P * P;
inline int mul(const int &a, const int &b) { return 1ll * a * b % P; }
inline int sub(int a, const int &b) {
  a -= b;
  return (a < 0) ? a + P : a;
}
inline int add(int a, const int &b) {
  a += b;
  return (a >= P) ? a - P : a;
}
int k;
long long n;
struct matrix {
  int num[31][31];
  matrix operator*(matrix a) {
    matrix c;
    for (int i = 0; i <= k; i++)
      for (int j = 0; j <= k; j++) {
        unsigned long long tem = 0;
        for (int l = 0; l <= k; l++)
          if ((tem += 1ll * num[i][l] * a.num[l][j]) >= p) tem -= p;
        c.num[i][j] = tem % P;
      }
    return c;
  }
} f[70][30];
int stk[70], top;
int main() {
  scanf("%lld%d", &n, &k);
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= k; j++) f[0][i].num[j][j] = f[0][i].num[j][i] = 1;
  }
  while (n) {
    stk[top++] = n % k;
    n /= k;
  }
  for (int i = 1; i < top; i++) {
    for (int j = 0; j < k; j++) {
      for (int l = 0; l <= k; l++) f[i][j].num[l][l] = 1;
      for (int l = 0; l < k; l++) f[i][j] = f[i][j] * f[i - 1][(j + l) % k];
    }
  }
  matrix ans;
  for (int l = 0; l <= k; l++)
    for (int m = 0; m <= k; m++) ans.num[l][m] = (l == k && m == k);
  for (int j = top - 1, sum = 0; ~j; --j) {
    for (int l = 0; l < stk[j]; l++) {
      ans = ans * f[j][sum];
      sum = (sum + 1) % k;
    }
  }
  int fin = 0;
  for (int i = 0; i <= k; i++) fin = add(fin, ans.num[k][i]);
  printf("%d\n", fin);
}