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;
struct Matrix {
  int n;
  long long a[50][50];
  Matrix() { memset(a, 0, sizeof(a)); }
};
Matrix operator*(Matrix A, Matrix B) {
  Matrix C;
  int m = C.n = A.n;
  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] + A.a[i][k] * B.a[k][j]) % mod;
  return C;
}
struct Vector {
  long long a[50];
  Vector() { memset(a, 0, sizeof(a)); }
};
Vector operator*(Matrix A, Vector x) {
  Vector y;
  int m = A.n;
  for (int i = 0; i < m; i++)
    for (int j = 0; j < m; j++) y.a[i] = (y.a[i] + A.a[i][j] * x.a[j]) % mod;
  return y;
}
long long get() {
  char c;
  while (c = getchar(), (c < '0' || c > '9') && (c != '-'))
    ;
  bool flag = (c == '-');
  if (flag) c = getchar();
  long long x = 0;
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - 48;
    c = getchar();
  }
  return flag ? -x : x;
}
void output(int x) {
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  int len = 0, data[10];
  while (x) {
    data[len++] = x % 10;
    x /= 10;
  }
  if (!len) data[len++] = 0;
  while (len--) putchar(data[len] + 48);
  putchar('\n');
}
Matrix f[60][50], l[50], r[50];
int main() {
  long long n = get();
  int k = get(), s = 0, len = 0;
  int d[60];
  while (n) {
    d[len++] = n % k;
    n /= k;
  }
  for (int i = 0; i < k; i++) {
    f[0][i].n = k + 1;
    for (int j = 0; j <= k; j++) f[0][i].a[i][j] = f[0][i].a[j][j] = 1;
  }
  for (int i = 1; i < len; i++) {
    l[0] = f[i - 1][0];
    for (int j = 1; j < k; j++) l[j] = f[i - 1][j] * l[j - 1];
    r[k - 1] = f[i - 1][k - 1];
    for (int j = k - 2; j >= 0; j--) r[j] = r[j + 1] * f[i - 1][j];
    f[i][0] = l[k - 1];
    for (int j = 1; j < k; j++) f[i][j] = l[j - 1] * r[j];
  }
  Vector x;
  x.a[k] = 1;
  for (int i = len - 1, s = 0; i >= 0; i--)
    for (int j = 0; j < d[i]; j++) {
      x = f[i][s] * x;
      s = (s + 1) % k;
    }
  long long ans = 0;
  for (int i = 0; i <= k; i++) ans += x.a[i];
  output(ans % mod);
  return 0;
}