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;
void nima(int &x, int md = MOD) {
  if (x >= md) x -= md;
  if (x < 0) x += md;
}
template <int N = 50>
struct Matrix {
  int n, m, M[N][N];
  Matrix(int _n = 0, int _m = 0, int x = 0) : n(_n), m(_m) {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) M[i][j] = x * (i == j);
  }
  int *operator[](int i) { return M[i]; }
  friend Matrix operator*(Matrix A, Matrix B) {
    assert(A.m == B.n);
    Matrix<N> C(A.n, B.m);
    for (int i = 0; i < A.n; i++)
      for (int k = 0; k < A.m; k++)
        for (int j = 0; j < B.m; j++)
          nima(C[i][j] += 1ll * A[i][k] * B[k][j] % MOD);
    return C;
  }
  void print() {
    for (int i = 0; i < n; i++, printf("\n"))
      for (int j = 0; j < m; j++) printf("%d ", M[i][j]);
  }
};
long long int n;
int k, X;
Matrix<> second(Matrix<> &A, int x) {
  Matrix<> ret(k + 1, k + 1);
  for (int i = 0; i < k; i++) {
    for (int j = 0; j < k; j++) ret[(i + x) % k][(j + x) % k] = A[i][j];
    ret[(i + x) % k][k] = A[i][k];
  }
  for (int i = 0; i <= k; i++) ret[k][i] = A[k][i];
  return ret;
}
int main() {
  scanf("%lld%d", &n, &k);
  vector<int> D;
  for (; n; n /= k) D.push_back(n % k), X += n % k;
  Matrix<> ret(k + 1, 1), M(k + 1, k + 1, 1);
  ret[k][0] = 1;
  for (int i = 0; i < k + 1; i++) M[0][i] = 1;
  for (int i = 0; i < (int)D.size(); i++) {
    nima(X -= D[i], k);
    for (int j = D[i] - 1; ~j; j--) ret = second(M, (j + X) % k) * ret;
    Matrix<> nM(k + 1, k + 1, 1);
    for (int j = 0; j < k; j++) nM = nM * second(M, j);
    swap(M, nM);
  }
  int ans = 0;
  for (int i = 0; i < k + 1; i++) nima(ans += ret[i][0]);
  printf("%d\n", ans);
  return 0;
}