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 long long mod = 1000000007;
struct mat {
  long long a[40][40];
  int n, m;
  mat() {
    memset(a, 0, sizeof(a));
    n = m = 0;
  }
  friend mat operator*(mat x, mat y) {
    mat res;
    res.n = x.n;
    res.m = y.m;
    for (int i = 0; i < x.n; i++) {
      for (int j = 0; j < y.m; j++) {
        for (int k = 0; k < x.m; k++) {
          res.a[i][j] += x.a[i][k] * y.a[k][j];
          res.a[i][j] %= mod;
        }
      }
    }
    return res;
  }
};
mat trans[70][40], pre[40], suc[40];
mat ans;
long long n, k;
int main() {
  cin >> n >> k;
  for (int i = 0; i < 40; i++) {
    for (int j = 0; j < 40; j++) {
      trans[i][j].n = k + 1;
      trans[i][j].m = k + 1;
    }
    pre[i].n = pre[i].m = k + 1;
    suc[i].n = suc[i].m = k + 1;
  }
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= k; j++) {
      trans[0][i].a[j][j] = 1;
    }
    for (int j = 0; j <= k; j++) {
      if (i != j) trans[0][i].a[i][j] = 1;
    }
  }
  ans.n = 1;
  ans.m = k + 1;
  for (int i = 0; i <= k; i++) ans.a[0][i] = 1;
  vector<int> rec;
  for (long long t = n; t; t /= k) rec.push_back(t % k);
  for (int i = 1; i < rec.size() + 3; i++) {
    pre[0] = trans[i - 1][0];
    for (int j = 1; j < k; j++) pre[j] = pre[j - 1] * trans[i - 1][j];
    suc[k - 1] = trans[i - 1][k - 1];
    for (int j = k - 2; j >= 0; j--) suc[j] = trans[i - 1][j] * suc[j + 1];
    trans[i][0] = suc[0];
    for (int j = 1; j < k; j++) trans[i][j] = suc[j] * pre[j - 1];
  }
  int d = 0;
  for (int i = rec.size() - 1; i >= 0; i--) {
    for (int j = 0; j < rec[i]; j++) {
      ans = ans * trans[i][(j + d) % k];
    }
    d = (d + rec[i]) % k;
  }
  cout << ans.a[0][k] << endl;
  return 0;
}