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;
template <typename T, typename TT>
ostream &operator<<(ostream &out, pair<T, TT> t) {
  return out << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream &operator<<(ostream &out, vector<T> t) {
  out << "[";
  for (auto x : t) out << x << " ";
  out << "]";
  return out;
}
template <typename T>
inline bool cmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool cmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T>
void read(T &x) {
  int f = 1;
  char ch;
  x = 0;
  while (!isdigit(ch = getchar()))
    if (ch == '-') f = -1;
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
const int mod = 1e9 + 7, KX = 32;
long long n;
int K;
struct matrix {
  int x[KX][KX];
  int *operator[](int p) { return x[p]; }
  matrix operator*(const matrix &b) {
    matrix ans = {};
    for (int i = 0; i <= K; i++) {
      for (int j = 0; j <= K; j++) {
        for (int k = 0; k <= K; k++) {
          ans[i][j] = (ans[i][j] + (long long)x[i][k] * b.x[k][j]) % mod;
        }
      }
    }
    return ans;
  }
} f[KX << 1][KX], p[KX], q[KX], ans;
int main() {
  vector<int> d;
  read(n);
  read(K);
  while (n) d.push_back(n % K), n /= K;
  for (int i = 0; i < K; i++) {
    for (int j = 0; j <= K; j++) {
      f[0][i][j][j] = f[0][i][j][i] = 1;
    }
  }
  ans[0][K] = 1;
  for (int i = 1; i < ((int)(d).size()); i++) {
    p[0] = f[i - 1][0];
    for (int j = 1; j < K; j++) p[j] = p[j - 1] * f[i - 1][j];
    q[K - 1] = f[i - 1][K - 1];
    for (int j = K - 2; j >= 0; j--) {
      q[j] = f[i - 1][j] * q[j + 1];
    }
    f[i][0] = q[0];
    for (int j = 1; j < K; j++) f[i][j] = q[j] * p[j - 1];
  }
  for (int i = ((int)(d).size()) - 1, l = 0; i >= 0; i--)
    for (int j = 0; j < d[i]; j++, l++) {
      ans = ans * f[i][l % K];
    }
  int result = 0;
  for (int i = 0; i <= K; i++) result = (result + ans[0][i]) % mod;
  cout << result << "\n";
  return 0;
}