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;
const int MODULO = (int)1e9 + 7;
vector<vector<int> > multiply(const vector<vector<int> > &a,
                              const vector<vector<int> > &b) {
  vector<vector<int> > c(a.size(), vector<int>(b[0].size()));
  for (int i = 0; i < (int)a.size(); ++i) {
    for (int j = 0; j < (int)a[0].size(); ++j) {
      for (int k = 0; k < (int)b[0].size(); ++k) {
        c[i][k] = (c[i][k] + 1LL * a[i][j] * b[j][k]) % MODULO;
      }
    }
  }
  return c;
}
int main() {
  ios::sync_with_stdio(false);
  long long n;
  cin >> n;
  int k;
  cin >> k;
  vector<vector<vector<int> > > a(k);
  for (int i = 0; i < k; ++i) {
    vector<vector<int> > &mat = a[i] =
        vector<vector<int> >(k + 1, vector<int>(k + 1));
    for (int j = 0; j < k; ++j)
      if (j != i) mat[j][j] = 1;
    mat[i][k] = 1;
    mat[k][k] = 2;
    mat[k][i] = MODULO - 1;
  }
  vector<vector<int> > mul(k + 1, vector<int>(k + 1));
  for (int i = 0; i <= k; ++i) mul[i][i] = 1;
  int sum = 0;
  for (long long m = n; m > 0; m /= k) sum += m % k;
  while (n > 0) {
    sum -= n % k;
    for (int i = n % k - 1; i >= 0; --i) mul = multiply(mul, a[(sum + i) % k]);
    vector<vector<vector<int> > > prefix(k);
    for (int i = 0; i < k; ++i) {
      if (i == 0)
        prefix[i] = a[i];
      else
        prefix[i] = multiply(a[i], prefix[i - 1]);
    }
    vector<vector<vector<int> > > suffix(k);
    for (int i = k - 1; i >= 0; --i) {
      if (i == k - 1)
        suffix[i] = a[i];
      else
        suffix[i] = multiply(suffix[i + 1], a[i]);
    }
    for (int i = 0; i < k; ++i) {
      if (i == 0)
        a[i] = suffix[i];
      else
        a[i] = multiply(prefix[i - 1], suffix[i]);
    }
    n /= k;
  }
  vector<vector<int> > res(k + 1, vector<int>(1));
  res[k][0] = 1;
  res = multiply(mul, res);
  cout << res[k][0] << '\n';
  return 0;
}