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 mod = 1000000007;
long long n;
int m;
struct Mat {
  int a[33][33];
  Mat() { memset(a, 0, sizeof(a)); }
} ans, A[63][33], pre[63][33], suf[63][33];
int bit[63];
inline Mat operator*(const Mat &a, const Mat &b) {
  Mat c;
  for (int i = 0; i <= m; i++)
    for (int j = 0; j <= m; j++)
      for (int k = 0; k <= m; k++)
        c.a[i][j] = (c.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;
  return c;
}
int main() {
  scanf("%I64d%d", &n, &m);
  long long tmp = n;
  int len = 0;
  while (tmp) bit[len++] = tmp % m, tmp /= m;
  for (int i = 0; i <= m; i++) ans.a[0][i] = 1;
  for (int i = 0; i < len; i++) {
    if (i == 0) {
      for (int j = 0; j < m; j++)
        for (int k = 0; k <= m; k++) A[0][j].a[k][k] = A[0][j].a[j][k] = 1;
    } else {
      A[i][0] = suf[i - 1][0];
      for (int j = 1; j < m; j++) A[i][j] = suf[i - 1][j] * pre[i - 1][j - 1];
    }
    pre[i][0] = A[i][0];
    for (int j = 1; j < m; j++) pre[i][j] = pre[i][j - 1] * A[i][j];
    suf[i][m - 1] = A[i][m - 1];
    for (int j = m - 2; j >= 0; j--) suf[i][j] = A[i][j] * suf[i][j + 1];
  }
  for (int i = len - 1, j = 0; i >= 0; i--)
    while (bit[i]--) ans = ans * A[i][j], j = (j + 1) % m;
  printf("%d", ans.a[0][m]);
  return 0;
}