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;
int mod = 1e9 + 7;
struct matrix {
  int A[40][40];
  int n;
  matrix(int x = 0, int _n = 0) {
    n = _n;
    memset(A, 0, sizeof A);
    for (int i = (0); i < (n); ++i) A[i][i] = x;
  }
  friend matrix operator*(const matrix &a, const matrix &b) {
    long long int mod2 = mod * 1LL * mod;
    matrix res(0, a.n);
    for (int i = (0); i < (a.n); ++i)
      for (int j = (0); j < (a.n); ++j) {
        long long int act = 0;
        for (int k = (0); k < (a.n); ++k) {
          act += a.A[i][k] * 1LL * b.A[k][j];
          if (act >= mod2) act -= mod2;
        }
        res.A[i][j] = act % mod;
      }
    return res;
  }
  friend vector<int> operator*(const matrix &a, const vector<int> &v) {
    vector<int> res;
    for (int i = (0); i < (a.n); ++i) {
      long long int act = 0;
      for (int k = (0); k < (a.n); ++k) {
        act += a.A[i][k] * 1LL * v[k];
        act %= mod;
      }
      res.push_back(act);
    }
    return res;
  }
};
void dbg(matrix m) {
  cerr << "macierz"
       << "=" << endl;
  for (int i = (0); i < (m.n); ++i) {
    for (int j = (0); j < (m.n); ++j) {
      cerr << m.A[i][j] << " ";
    }
    cerr << endl;
  }
}
matrix ch[100][50];
matrix pref[50], suf[50];
int main() {
  long long int n;
  int k;
  cin >> n >> k;
  n--;
  vector<int> digits;
  int l = 0;
  while (n > 0) {
    digits.push_back(n % k);
    n = n / k;
    l++;
  }
  for (int i = (0); i < (k); ++i) {
    ch[0][i] = matrix(1, k + 1);
    for (int j = (0); j < (k + 1); ++j) ch[0][i].A[i][j] = 1;
  }
  for (int i = (1); i < (l + 1); ++i) {
    pref[0] = ch[i - 1][0];
    for (int j = (1); j < (k); ++j) pref[j] = ch[i - 1][j] * pref[j - 1];
    suf[k - 1] = ch[i - 1][k - 1];
    for (int j = (k - 2); j >= (0); --j) {
      suf[j] = suf[j + 1] * ch[i - 1][j];
    }
    ch[i][0] = suf[0];
    for (int j = (1); j < (k); ++j) {
      ch[i][j] = pref[j - 1] * suf[j];
    }
  }
  matrix m = matrix(1, k + 1);
  int reszta = 0;
  for (int i = (l - 1); i >= (0); --i) {
    for (int j = (0); j < (digits[i]); ++j) {
      m = ch[i][(reszta + j) % k] * m;
    }
    reszta = (reszta + digits[i]) % k;
  }
  m = ch[0][reszta] * m;
  vector<int> w;
  for (int i = (0); i < (k); ++i) w.push_back(0);
  w.push_back(1);
  w = m * w;
  long long int ans = 0;
  for (int i = (0); i < (w.size()); ++i) ans = (ans + w[i]) % mod;
  cout << ans << endl;
  return 0;
}