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 MAXK = 30;
const int mod = 1000000007;
int add(int a, int b) { return (a + b) % mod; }
int mul(int a, int b) { return (long long)a * b % mod; }
long long N;
int K;
struct Matrix {
  int m[MAXK + 1][MAXK + 1];
};
void gen_id(Matrix &M) {
  for (int i = (0); i < (K + 1); ++i)
    for (int j = (0); j < (K + 1); ++j) M.m[i][j] = 0;
  for (int i = (0); i < (K + 1); ++i) M.m[i][i] = 1;
}
void gen_matrix(Matrix &M, int r) {
  gen_id(M);
  for (int i = (0); i < (K + 1); ++i) M.m[i][r] = 1;
}
void output(Matrix &M) {
  for (int i = (0); i < (K + 1); ++i) {
    for (int j = (0); j < (K + 1); ++j) printf("%d ", M.m[i][j]);
    puts("");
  }
}
Matrix tmp;
void mule(Matrix &a, const Matrix &b) {
  for (int i = (0); i < (K + 1); ++i)
    for (int j = (0); j < (K + 1); ++j) tmp.m[i][j] = 0;
  for (int i = (0); i < (K + 1); ++i)
    for (int j = (0); j < (K + 1); ++j)
      for (int k = (0); k < (K + 1); ++k)
        tmp.m[i][j] = add(tmp.m[i][j], mul(a.m[i][k], b.m[k][j]));
  a = tmp;
}
void rot(Matrix &M, int r) {
  for (int i = (0); i < (K + 1); ++i)
    for (int j = (0); j < (K); ++j) tmp.m[i][j] = M.m[i][(j + r) % K];
  for (int i = (0); i < (K + 1); ++i) tmp.m[i][K] = M.m[i][K];
  M = tmp;
  for (int j = (0); j < (K); ++j)
    for (int i = (0); i < (K); ++i) tmp.m[i][j] = M.m[(i + r) % K][j];
  M = tmp;
}
const int MAXPOS = 61;
int len, x[MAXPOS];
Matrix memo[MAXPOS];
int bio[MAXPOS];
Matrix solve(bool dira, int pos) {
  if (pos == len) {
    Matrix ret;
    if (dira)
      gen_id(ret);
    else
      gen_matrix(ret, 0);
    return ret;
  }
  if (!dira && bio[pos]) return memo[pos];
  Matrix ret;
  gen_id(ret);
  int lo = 0, hi = dira ? x[pos] : K - 1;
  for (int d = lo; d <= hi; ++d) {
    Matrix now = solve(dira && d == hi, pos + 1);
    rot(now, d);
    mule(ret, now);
  }
  if (!dira) {
    bio[pos] = true;
    memo[pos] = ret;
  }
  return ret;
}
int main(void) {
  scanf("%I64d%d", &N, &K);
  len = 0;
  while (N) {
    x[len] = N % K;
    N /= K;
    ++len;
  }
  reverse(x, x + len);
  int sol = 0;
  Matrix ret = solve(true, 0);
  for (int i = (0); i < (K + 1); ++i) sol = add(sol, ret.m[K][i]);
  printf("%d\n", sol);
  return 0;
}