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 N = 31;
const int Q = 1e9 + 7;
struct Matrix {
  int n, m;
  long long a[N][N];
  Matrix() {
    n = m = 0;
    memset(a, 0, sizeof(a));
  }
  Matrix(int _n, int _m) {
    n = _n, m = _m;
    memset(a, 0, sizeof(a));
  }
  long long* operator[](int x) { return a[x]; }
  const long long* operator[](int x) const { return a[x]; }
};
Matrix operator*(const Matrix& A, const Matrix& B) {
  Matrix res(A.n, B.m);
  for (int i = 0; i < A.n; ++i)
    for (int j = 0; j < A.m; ++j)
      for (int k = 0; k < B.m; ++k) {
        res[i][k] += A[i][j] * B[j][k] % Q;
        res[i][k] %= Q;
      }
  return res;
}
long long n;
int K;
int dig[60], len;
Matrix M[60][30], pre[30], suf[30];
void work() {
  cin >> n >> K;
  int i, j, k;
  while (n) {
    dig[len++] = n % K;
    n /= K;
  }
  for (i = 0; i < K; ++i) {
    M[0][i].n = M[0][i].m = K + 1;
    for (j = 0; j <= K; ++j) {
      M[0][i][j][j] = 1;
      M[0][i][j][i] = 1;
    }
  }
  for (k = 1; k < len; ++k) {
    pre[0] = M[k - 1][0];
    for (i = 1; i < K; ++i) pre[i] = pre[i - 1] * M[k - 1][i];
    suf[K - 1] = M[k - 1][K - 1];
    for (i = K - 2; i >= 0; --i) suf[i] = M[k - 1][i] * suf[i + 1];
    for (i = 0; i < K; ++i) {
      M[k][i] = suf[i];
      if (i) M[k][i] = M[k][i] * pre[i - 1];
    }
  }
  Matrix I(1, K + 1);
  I[0][K] = 1;
  int x = 0;
  for (k = len - 1; k >= 0; --k) {
    for (j = 0; j < dig[k]; ++j) {
      I = I * M[k][(x + j) % K];
    }
    x += dig[k], x %= K;
  }
  long long res = 0;
  for (i = 0; i <= K; ++i) res += I[0][i], res %= Q;
  cout << res << endl;
}
int main() {
  work();
  return 0;
}