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 mo = 1e9 + 7;
const double pi = acos(-1);
long long N;
int K, n, a[120];
struct Matrix {
  int a[32][32];
  int* operator[](int x) { return a[x]; }
  friend Matrix operator*(Matrix a, Matrix b) {
    Matrix c;
    memset(&c, 0, sizeof(c));
    for (int i = 0; i <= K; ++i)
      for (int j = 0; j <= K; ++j)
        for (int k = 0; k <= K; ++k)
          c[i][k] = (c[i][k] + 1ll * a[i][j] * b[j][k]) % mo;
    return c;
  }
} f[120][32], ans, suc[32], pre[32];
Matrix I() {
  Matrix tmp;
  memset(&tmp, 0, sizeof(tmp));
  for (int i = 0; i <= K; ++i) tmp[i][i] = 1;
  return tmp;
}
void prep() {
  for (int i = 0; i <= K - 1; ++i) {
    for (int j = 0; j <= K; ++j) f[0][i][j][j] = 1;
    for (int j = 0; j <= K; ++j)
      if (i != j) f[0][i][i][j] = 1;
  }
  for (int i = 1; i <= n; ++i) {
    pre[0] = f[i - 1][0];
    for (int j = 1; j <= K - 1; ++j) pre[j] = pre[j - 1] * f[i - 1][j];
    suc[K - 1] = f[i - 1][K - 1];
    for (int j = K - 2; j >= 0; --j) suc[j] = f[i - 1][j] * suc[j + 1];
    for (int j = 0; j <= K - 1; ++j)
      f[i][j] = suc[j] * (j == 0 ? I() : pre[j - 1]);
  }
}
int main() {
  scanf("%I64d%d", &N, &K);
  while (N) {
    a[++n] = N % K;
    N /= K;
  }
  prep();
  for (int i = 0; i <= K; ++i) ans[1][i] = 1;
  int x = 0;
  for (int i = n; i >= 1; --i) {
    for (int j = 0; j <= a[i] - 1; ++j) ans = ans * f[i - 1][(j + x) % K];
    x = (x + a[i]) % K;
  }
  printf("%d\n", ans[1][K]);
}