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 N = 100, D = 31, MOD = 1E9 + 7;
int d, cur, a[N];
long long n;
struct matrix {
  long long f[D][D];
  matrix() {
    for (int i = 0; i <= d; i++) {
      for (int j = 0; j <= d; j++) {
        f[i][j] = 0;
      }
    }
  }
  matrix operator*(const matrix& oth) const {
    matrix ans;
    for (int i = 0; i <= d; i++) {
      for (int j = 0; j <= d; j++) {
        for (int k = 0; k <= d; k++) {
          (ans.f[i][j] += f[i][k] * oth.f[k][j]) %= MOD;
        }
      }
    }
    return ans;
  }
} bas[N];
matrix rotate(int u) {
  (u += d) %= d;
  matrix ans;
  for (int i = 0; i < d; i++) {
    ans.f[i][(i + u) % d] = 1;
  }
  ans.f[d][d] = 1;
  return ans;
}
void init() {
  for (int i = 1; i < d; i++) {
    bas[0].f[i][i] = 1;
  }
  bas[0].f[d][0] = 1;
  bas[0].f[d][d] = 2;
  bas[0].f[0][d] = -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n >> d;
  for (cur = 0; n > 0; cur++, n /= d) {
    a[cur] = n % d;
  }
  init();
  for (int i = 1; i < cur; i++) {
    bas[i] = rotate(0);
    for (int j = 0; j < d; j++) {
      bas[i] = bas[i] * bas[i - 1] * rotate(1);
    }
  }
  matrix ans = rotate(0);
  for (int i = cur - 1; i >= 0; i--) {
    for (int j = 0; j < a[i]; j++) {
      ans = ans * bas[i] * rotate(1);
    }
  }
  cout << (ans.f[d][d] + MOD) % MOD;
}