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 double eps(1e-8);
const double pi(3.14159265358979);
const int K = 33, mod = 1000000007;
struct matrix {
  long long a[K][K];
} t[K] = {}, ans = {{}}, pre[K] = {}, suf[K] = {};
long long n;
int k;
inline void multi(const matrix &m1, const matrix &m2, matrix &to) {
  static matrix tmp = {{}};
  for (int i = 0; i <= k; ++i)
    for (int j = 0; j <= k; ++j) {
      tmp.a[i][j] = 0;
      for (int p = 0; p <= k; ++p)
        (tmp.a[i][j] += m1.a[i][p] * m2.a[p][j]) %= mod;
    }
  to = tmp;
}
void init() {
  cin >> n >> k;
  for (int i = 0; i <= k; ++i)
    for (int j = 0; j <= k; ++j) ans.a[i][j] = i == j;
  for (int p = 0; p < k; ++p)
    for (int i = 0; i <= k; ++i)
      for (int j = 0; j <= k; ++j) t[p].a[i][j] = i == j || i == p;
}
void work() {
  vector<int> bit;
  for (; n; n /= k) bit.push_back(n % k);
  bit.push_back(0);
  partial_sum(bit.rbegin(), bit.rend(), bit.rbegin());
  for (auto p = 0u; p < bit.size() - 1; ++p) {
    for (int i = bit[p] - 1; i >= bit[p + 1]; --i) multi(t[i % k], ans, ans);
    pre[0] = t[0];
    for (int i = 1; i < k; ++i) multi(pre[i - 1], t[i], pre[i]);
    suf[k - 1] = t[k - 1];
    for (int i = k - 2; i >= 0; --i) multi(t[i], suf[i + 1], suf[i]);
    t[0] = suf[0];
    for (int i = 1; i < k; ++i) multi(suf[i], pre[i - 1], t[i]);
  }
  long long sum = 0;
  for (int i = 0; i <= k; ++i) sum += ans.a[i][k];
  cout << sum % mod << endl;
}
int main() {
  init();
  work();
  return 0;
}