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 MAXL = 100;
const int MAXK = 31;
const int MOD = 1000000007;
struct matrix {
  int n, m;
  int val[MAXK][MAXK];
  matrix(int n = 0, int m = 0) : n(n), m(m) { memset(val, 0, sizeof val); }
  int operator()(int i, int j) const { return val[i][j]; }
  int &operator()(int i, int j) { return val[i][j]; }
};
matrix operator*(const matrix &A, const matrix &B) {
  matrix C(A.n, B.m);
  for (int i = 0; i < A.n; ++i)
    for (int j = 0; j < B.m; ++j)
      for (int k = 0; k < A.m; ++k)
        C(i, j) = (C(i, j) + (long long)A(i, k) * B(k, j)) % MOD;
  return C;
}
long long n;
int k, len;
int dig[MAXL];
matrix M[MAXL][MAXK];
matrix L[MAXL][MAXK];
matrix R[MAXL][MAXK];
void print(const matrix &A) {
  for (int i = 0; i < A.n; ++i) {
    for (int j = 0; j < A.n; ++j) printf("%d ", A(i, j));
    printf("\n");
  }
  printf("----------------------\n");
}
int main(void) {
  scanf("%lld%d", &n, &k);
  matrix I(k + 1, k + 1);
  for (int i = 0; i < k + 1; ++i) I(i, i) = 1;
  for (int d = 0; d < k; ++d) {
    M[0][d] = I;
    for (int j = 0; j <= k; ++j) M[0][d](j, d) = 1;
  }
  int m = 0;
  for (long long tmp = n; tmp > 0; tmp /= k) {
    if (m >= 1) {
      for (int d = 0; d < k; ++d) M[m][d] = R[m - 1][d] * L[m - 1][d];
    }
    L[m][0] = I;
    for (int d = 1; d < k; ++d) L[m][d] = L[m][d - 1] * M[m][d - 1];
    R[m][k - 1] = M[m][k - 1];
    for (int d = k - 2; d >= 0; --d) R[m][d] = M[m][d] * R[m][d + 1];
    dig[m++] = tmp % k;
  }
  reverse(dig, dig + m);
  matrix ans = I;
  int d = 0;
  for (int i = 0; i < m; ++i)
    for (int p = 0; p < dig[i]; ++p) {
      d = (d + 1) % k;
      ans = ans * M[m - i - 1][d];
    }
  int sol = 0;
  for (int i = 0; i < k + 1; ++i) sol = (sol + ans(k, i)) % MOD;
  printf("%d\n", sol);
  return 0;
}