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 = 1e9 + 7;
const int MAXN = 35;
int tmp[MAXN], K;
struct Matrix {
  int a[MAXN][MAXN];
  Matrix trans(int x) {
    Matrix c;
    for (int i = 0; i < K; ++i) tmp[i] = (i + x) % K;
    tmp[K] = K;
    for (int i = 0; i <= K; ++i)
      for (int j = 0; j <= K; ++j) c.a[tmp[i]][tmp[j]] = a[i][j];
    return c;
  }
  Matrix operator*(const Matrix b) const {
    Matrix res;
    for (int i = 0; i <= K; ++i)
      for (int j = 0; j <= K; ++j) res.a[i][j] = 0;
    for (int i = 0; i <= K; ++i)
      for (int j = 0; j <= K; ++j)
        for (int l = 0; l <= K; ++l)
          res.a[i][j] = (res.a[i][j] + 1LL * a[i][l] * b.a[l][j]) % mod;
    return res;
  }
  void prt() {
    for (int i = 0; i <= K; ++i, printf("\n"))
      for (int j = 0; j <= K; ++j) printf("%d ", a[i][j]);
    puts("");
  }
} f[70][MAXN], ans;
long long n;
void init() {
  cin >> n >> K;
  for (int i = 0; i <= K; ++i)
    ans.a[i][i] = f[0][0].a[i][i] = f[0][0].a[i][0] = 1;
}
void work() {
  int T(0);
  long long pow(1);
  for (;;) {
    for (int i = 1; i < K; ++i) f[T][i] = f[T][i - 1] * (f[T][0].trans(i));
    f[T + 1][0] = f[T][K - 1];
    if (pow > n / K) break;
    ++T;
    pow *= 1LL * K;
  }
  int add(0);
  for (; T >= 0; --T, pow /= 1LL * K) {
    if (n / pow) ans = ans * (f[T][n / pow - 1].trans(add));
    add = (add + n / pow) % K;
    n %= pow;
  }
  long long sum(0);
  for (int i = 0; i <= K; ++i) sum += ans.a[K][i];
  cout << (sum % mod) << endl;
}
int main() {
  init();
  work();
  return 0;
}