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 int modulo = 1e9 + 7;
int n;
vector<vector<long long int> > cero, identidad;
vector<vector<long long int> > operator*(vector<vector<long long int> > m1,
                                         vector<vector<long long int> > m2) {
  vector<vector<long long int> > m(n, vector<long long int>(n, 0));
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++)
        m[i][j] = (m[i][j] + m1[i][k] * m2[k][j]) % modulo;
  return m;
}
vector<long long int> operator*(vector<vector<long long int> > m,
                                vector<long long int> v) {
  vector<long long int> r(n, 0);
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) r[i] = (r[i] + m[i][j] * v[j]) % modulo;
  return r;
}
vector<vector<long long int> > s[50];
int main() {
  long long int x;
  int b;
  cin >> x >> b;
  x--;
  n = b + 1;
  identidad = cero =
      vector<vector<long long int> >(n, vector<long long int>(n, 0));
  for (int i = 0; i < n; i++) identidad[i][i] = 1;
  s[0] = identidad;
  s[1] = cero;
  for (int i = 0; i < n - 2; i++) s[1][i][i + 1] = 1;
  s[1][n - 2][0] = 1;
  s[1][n - 1][n - 1] = 1;
  for (int i = 2; i <= b - 1; i++) s[i] = s[1] * s[i - 1];
  vector<vector<long long int> > N = identidad;
  N[0][0] = 0;
  N[0][n - 1] = 1;
  N[n - 1][0] = -1;
  N[n - 1][n - 1] = 2;
  vector<vector<long long int> > M = N;
  while (x) {
    int d = x % b;
    x /= b;
    vector<vector<long long int> > nextM = M;
    for (int i = 1; i < b; i++) {
      if (i == d) N = s[b - i] * N * s[i] * nextM;
      nextM = s[b - i] * M * s[i] * nextM;
    }
    M = nextM;
  }
  vector<long long int> v(n, 0);
  v[n - 1] = 1;
  long long int sol = (N * v)[n - 1];
  cout << (sol + modulo) % modulo << endl;
}