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 inf = int(1e9);
const double eps = 1e-9;
const double pi = 4 * atan(double(1));
const int K = 31;
const int MOD = int(1e9) + 7;
int k = 0;
struct matr {
  int a[K][K];
  matr() {
    for (int i = 0; i <= k; ++i) {
      for (int j = 0; j <= k; ++j) {
        a[i][j] = 0;
      }
      a[i][i] = 1;
    }
  }
};
matr operator*(const matr &a, const matr &b) {
  matr c;
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= k; ++j) {
      c.a[i][j] = 0;
      for (int z = 0; z <= k; ++z) {
        c.a[i][j] = (c.a[i][j] + 1LL * a.a[i][z] * b.a[z][j]) % MOD;
      }
    }
  }
  return c;
}
matr M[K], NM[K], suff[K], pref[K];
inline matr get(int num) {
  matr res;
  for (int i = 0; i <= k; ++i) {
    res.a[num][i] = 1;
  }
  return res;
}
int main() {
  long long n;
  cin >> n >> k;
  vector<int> v;
  while (n > 0) {
    v.push_back(n % k);
    n /= k;
  }
  matr res;
  int shift = 0;
  for (int i = 0; i < int((v).size()); ++i) {
    shift += v[i];
  }
  for (int i = 0; i < k; ++i) {
    M[i] = matr();
    pref[i] = matr();
    suff[i] = matr();
  }
  for (int i = 0; i < int((v).size()); ++i) {
    shift -= v[i];
    for (int j = v[i] - 1; j >= 0; --j) {
      if (i == 0) {
        res = get((shift + j) % k) * res;
        continue;
      }
      res = M[(shift + j) % k] * res;
    }
    if (i == 0) {
      for (int i = 0; i < k; ++i) {
        M[i] = matr();
        for (int j = k - 1; j >= 0; --j) {
          M[i] = get((i + j) % k) * M[i];
        }
      }
    } else {
      for (int i = 0; i < k; ++i) {
        M[i] = matr();
        if (i > 0) {
          M[i] = pref[i - 1] * M[i];
        }
        M[i] = suff[i] * M[i];
      }
    }
    for (int i = 0; i < k; ++i) {
      pref[i] = (i > 0 ? pref[i - 1] : matr()) * M[i];
    }
    for (int i = k - 1; i >= 0; --i) {
      suff[i] = M[i] * (i + 1 < k ? suff[i + 1] : matr());
    }
  }
  int ans = 0;
  for (int i = 0; i <= k; ++i) {
    ans = (ans + res.a[i][k]) % MOD;
  }
  cout << ans << endl;
  return 0;
}