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 k;
struct mat {
  long long m[34][34];
};
mat ch(mat a, mat b) {
  mat c;
  memset(c.m, 0, sizeof(c.m));
  for (int i = 0; i <= k; i++)
    for (int u = 0; u <= k; u++)
      for (int j = 0; j <= k; j++)
        c.m[i][j] = (c.m[i][j] + a.m[i][u] * b.m[u][j]) % 1000000007;
  return c;
}
mat ks(mat a, long long b) {
  mat sum;
  memset(sum.m, 0, sizeof(sum.m));
  for (int i = 0; i <= k; i++) sum.m[i][i] = 1;
  while (b) {
    if (b & 1) sum = ch(sum, a);
    a = ch(a, a);
    b <<= 2;
  }
  return sum;
}
mat zz[70][33], ans, pr[33], la[33];
int q[70], all;
int main() {
  scanf("%I64d%d", &n, &k);
  n--;
  for (int i = 0; i <= k; i++) ans.m[i][i] = 1;
  for (int i = 0; i <= k; i++) pr[0].m[i][i] = la[k].m[i][i] = 1;
  for (int i = 0; i < k; i++) {
    for (int j = 0; j <= k; j++) {
      zz[0][i].m[j][j] = 1;
      zz[0][i].m[i][j] = 1;
    }
  }
  while (n) {
    q[all++] = n % k;
    n /= k;
  }
  for (int i = 1; i < all; i++) {
    for (int j = 0; j < k; j++) pr[j + 1] = ch(pr[j], zz[i - 1][j]);
    for (int j = k - 1; j >= 0; j--) la[j] = ch(zz[i - 1][j], la[j + 1]);
    for (int j = 0; j < k; j++) zz[i][j] = ch(la[j], pr[j]);
  }
  int now = 0;
  for (int i = all - 1; i >= 0; i--) {
    for (int j = 0; j < q[i]; j++) ans = ch(ans, zz[i][(now + j) % k]);
    now = (now + q[i]) % k;
  }
  ans = ch(ans, zz[0][now]);
  long long ret = 0;
  for (int i = 0; i <= k; i++) ret = (ret + ans.m[i][k]) % 1000000007;
  cout << ret << endl;
}