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 n;
int m, f[65][35][35], s[65], num, g[35][35], A[35], B[35];
inline int Dec(int x, int y) { return x < m ? (x - y + m) % m : m; }
int main() {
  cin >> n >> m;
  for (long long t = n; t; t /= m) s[++num] = t % m;
  for (int i = 1; i <= m; ++i) f[1][i][i] = 1;
  for (int i = 0; i <= m; ++i) f[1][0][i] = 1;
  for (int i = 2; i <= num; ++i) {
    for (int j = 0; j <= m; ++j) f[i][j][j] = 1;
    for (int j = 0; j < m; ++j) {
      for (int a = 0; a <= m; ++a)
        for (int b = 0; b <= m; ++b)
          for (int c = 0; c <= m; ++c)
            g[a][c] =
                (g[a][c] + 1LL * f[i - 1][Dec(a, j)][Dec(b, j)] * f[i][b][c]) %
                1000000007;
      for (int a = 0; a <= m; ++a)
        for (int b = 0; b <= m; ++b) f[i][a][b] = g[a][b], g[a][b] = 0;
    }
  }
  for (int i = 0; i <= m; ++i) A[i] = 1;
  for (int i = 1; i <= num; ++i) {
    int S = 0;
    for (int j = i + 1; j <= num; ++j) S = (S + s[j]) % m;
    for (; s[i]; --s[i]) {
      int dec = (S + s[i]) % m;
      for (int a = 0; a <= m; ++a)
        for (int b = 0; b <= m; ++b)
          B[b] =
              (B[b] + 1LL * A[a] * f[i][Dec(a, dec)][Dec(b, dec)]) % 1000000007;
      for (int a = 0; a <= m; ++a) A[a] = B[a], B[a] = 0;
    }
  }
  printf("%d\n", A[m]);
  return 0;
}