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;
long long bin[33 * 2], n, m;
struct matrix {
  long long mat[33][33];
  inline matrix operator*(const matrix &a) const {
    matrix res;
    memset(res.mat, 0, sizeof(res.mat));
    for (long long i = 0; i <= m; ++i) {
      for (long long j = 0; j <= m; ++j) {
        for (long long k = 0; k <= m; ++k) {
          res.mat[i][j] =
              (res.mat[i][j] + mat[i][k] * a.mat[k][j]) % 1000000007;
        }
      }
    }
    return res;
  }
} ans, P[33 * 2][33], S[33 * 2][33], A[33 * 2][33];
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
signed main() {
  n = read(), m = read();
  long long n2 = n, len = -1;
  while (n2) {
    len++;
    bin[len] = n2 % m;
    n2 = n2 / m;
  }
  len++;
  for (long long i = 0; i <= m; ++i) ans.mat[0][i] = 1;
  for (long long i = 0; i < len; ++i) {
    for (long long j = 0; j <= m; ++j) A[i][0].mat[j][j] = 1;
    if (!i) {
      for (long long j = 0; j < m; ++j) {
        A[i][j].mat[m][m] = 1;
        for (long long k = 0; k < m; ++k) {
          if (k != j) {
            A[i][j].mat[k][k] = 1;
            continue;
          }
          for (long long l = 0; l <= m; ++l) A[i][j].mat[k][l] = 1;
        }
      }
    } else {
      for (long long j = 0; j < m; ++j) {
        if (!j)
          A[i][j] = S[i - 1][0];
        else
          A[i][j] = S[i - 1][j] * P[i - 1][j - 1];
      }
    }
    P[i][0] = A[i][0];
    for (long long j = 1; j < m; ++j) P[i][j] = P[i][j - 1] * A[i][j];
    S[i][m - 1] = A[i][m - 1];
    for (long long j = m - 2; j >= 0; --j) S[i][j] = A[i][j] * S[i][j + 1];
  }
  long long j = 0;
  for (long long i = len - 1; i >= 0; --i) {
    while (bin[i]-- > 0) {
      ans = ans * A[i][j];
      j++;
      if (j == m) j = 0;
    }
  }
  cout << ans.mat[0][m] << endl;
}