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;
int m;
struct M {
  int a[35][35];
  M() { memset(a, 0, sizeof(a)); }
};
M I() {
  M ans;
  for (int i = 0; i <= m; i++) ans.a[i][i] = 1;
  return ans;
}
M operator*(const M& a, const M& b) {
  M ans;
  for (int i = 0; i <= m; i++)
    for (int j = 0; j <= m; j++)
      for (int k = 0; k <= m; k++)
        ans.a[i][j] = (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j]) % Mod;
  return ans;
}
long long n;
M f[65][35], g[35], h[35];
int a[65];
int main() {
  cin >> n >> m;
  int N = 0;
  while (n) {
    a[N++] = n % m;
    n /= m;
  }
  N--;
  for (int i = 0; i < m; i++) {
    f[0][i] = I();
    for (int j = 0; j <= m; j++) f[0][i].a[j][i] = 1;
  }
  for (int i = 1; i <= N; i++) {
    g[0] = h[m] = I();
    for (int j = 1; j < m; j++) g[j] = g[j - 1] * f[i - 1][j - 1];
    for (int j = m - 1; j >= 0; j--) h[j] = f[i - 1][j] * h[j + 1];
    for (int j = 0; j < m; j++) f[i][j] = h[j] * g[j];
  }
  M s;
  s.a[0][m] = 1;
  int cur = 0;
  for (int i = N; i >= 0; i--)
    for (int j = 0; j < a[i]; j++) {
      s = s * f[i][cur];
      cur = (cur + 1) % m;
    }
  int ans = 0;
  for (int i = 0; i <= m; i++) ans = (ans + s.a[0][i]) % Mod;
  printf("%d\n", ans);
  return 0;
}