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 = 1000000007;
inline int add(int a, int b) {
  if ((a += b) >= mod) a -= mod;
  return a;
}
inline int mult(int a, int b) {
  long long t = 1ll * a * b;
  if (t >= mod) t %= mod;
  return t;
}
namespace matrix {
template <int N, int M>
class matr {
 public:
  int dat[N][M];
  void init() { memset(dat, 0, sizeof(dat)); }
  void iden() {
    init();
    for (int i = 0; i < min(N, M); i++) dat[i][i] = 1;
  }
  matr<M, N> trans() {
    matr<M, N> out;
    for (int i = 0; i < M; i++)
      for (int j = 0; j < N; j++) out[i][j] = dat[j][i];
  }
  const int *operator[](const int &x) const { return dat[x]; }
  int *operator[](const int &x) { return dat[x]; }
  matr<N, M> operator+(const matr<N, M> a) {
    matr<N, M> out;
    for (int i = 0; i < N; i++)
      for (int j = 0; j < M; j++) out[i][j] = add(dat[i][j], a[i][j]);
    return out;
  }
  matr<N, M> operator*(const int &x) {
    matr<N, M> out = *this;
    for (int i = 0; i < N; i++)
      for (int j = 0; j < M; j++) out[i][j] = mult(out[i][j], x);
    return out;
  }
  template <int X>
  friend matr<N, M> operator*(const matr<N, X> a, const matr<X, M> b) {
    matr<N, M> out;
    out.init();
    for (int i = 0; i < N; i++) {
      for (int j = 0; j < X; j++) {
        if (a[i][j]) {
          for (int k = 0; k < M; k++)
            out[i][k] = add(out[i][k], mult(a[i][j], b[j][k]));
        }
      }
    }
    return out;
  }
};
};  // namespace matrix
using namespace matrix;
long long n;
int K;
vector<int> digit;
matr<35, 35> A[65][35], S[65][35], P[65][35], ans;
int main() {
  scanf("%lld%d", &n, &K);
  for (int i = 0; i < K; i++) {
    A[0][i].iden();
    for (int j = 0; j <= K; j++) A[0][i][i][j] = 1;
  }
  long long len = K;
  for (int i = 1; len <= n; i++, len *= K) {
    S[i - 1][K - 1] = A[i - 1][K - 1];
    for (int j = K - 2; j >= 0; j--)
      S[i - 1][j] = A[i - 1][j] * S[i - 1][j + 1];
    P[i - 1][0] = A[i - 1][0];
    for (int j = 1; j < K; j++) P[i - 1][j] = P[i - 1][j - 1] * A[i - 1][j];
    A[i][0] = S[i - 1][0];
    for (int j = 1; j < K; j++) A[i][j] = S[i - 1][j] * P[i - 1][j - 1];
  }
  while (n) {
    digit.push_back(n % K);
    n /= K;
  }
  int s = 0;
  ans.iden();
  for (int i = digit.size() - 1; i >= 0; i--) {
    for (int j = 0; j < digit[i]; j++) ans = ans * A[i][(j + s) % K];
    s = (s + digit[i]) % K;
  }
  int res = 1;
  for (int i = 0; i < K; i++) res = add(res, ans[i][K]);
  printf("%d\n", res);
  return 0;
}