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>
inline int read() {
  int ret;
  scanf("%d", &ret);
  return ret;
}
int dx8[8] = {1, 1, 0, -1, -1, -1, 0, 1};
int dy8[8] = {0, 1, 1, 1, 0, -1, -1, -1};
int dx4[4] = {1, 0, -1, 0};
int dy4[4] = {0, 1, 0, -1};
using namespace std;
inline void add(long long &x, long long y) { x = (x + y) % 1000000007ll; }
int r;
struct Matrix {
  long long a[33][33];
  void toZero() { memset(a, 0, sizeof a); }
  void toE() {
    toZero();
    for (int i = 0; i < int(r); i++) a[i][i] = 1;
  }
  friend Matrix operator*(const Matrix &a, const Matrix &b) {
    Matrix ret;
    ret.toZero();
    for (int i = 0; i < int(r); i++)
      for (int j = 0; j < int(r); j++)
        for (int k = 0; k < int(r); k++)
          add(ret.a[i][j], a.a[i][k] * b.a[k][j]);
    return ret;
  }
};
long long n, K;
Matrix one[70];
Matrix two[70];
Matrix rotr[70], rotc[70];
int Main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> K;
  r = K + 1;
  for (int i = 0; i < int(K + 1); i++) {
    one[i].toE();
    if (i)
      for (int j = 0; j < int(K + 1); j++) one[i].a[i][j] = 1;
  }
  for (int i = 0; i < int(K + 1); i++) {
    rotr[i].toE();
    if (i == 1) {
      rotr[i].toZero();
      rotr[i].a[0][0] = 1;
      rotr[i].a[1][K] = 1;
      for (int j = 2; j <= K; j++) rotr[i].a[j][j - 1] = 1;
    } else if (i > 1)
      rotr[i] = rotr[1] * rotr[i - 1];
  }
  for (int i = 0; i < int(K + 1); i++) {
    rotc[i].toE();
    if (i == 1) {
      rotc[i].toZero();
      rotc[i].a[0][0] = 1;
      rotc[i].a[K][1] = 1;
      for (int j = 2; j <= K; j++) rotc[i].a[j - 1][j] = 1;
    } else if (i > 1)
      rotc[i] = rotc[i - 1] * rotc[1];
  }
  two[1].toE();
  for (int i = 1; i <= int(K); i++) two[1] = one[i] * two[1];
  int bit = 0;
  int num[70];
  while (n) {
    num[bit++] = n % K;
    n /= K;
  }
  for (int i = 2; i <= bit; i++) {
    two[i].toE();
    for (int j = 0; j < int(K); j++)
      two[i] = rotr[j] * two[i - 1] * rotc[j] * two[i];
  }
  reverse(num, num + bit);
  Matrix ret;
  ret.toE();
  int ps = 0;
  for (int i = 0; i < int(bit); i++) {
    if (i == bit - 1)
      for (int j = 0; j < int(num[i]); j++) ret = one[ps % K + 1] * ret, ps++;
    else
      for (int j = 0; j < int(num[i]); j++)
        ret = rotr[ps % K] * two[bit - i - 1] * rotc[ps % K] * ret, ps++;
  }
  long long res = 1;
  for (int i = 1; i <= int(K); i++) add(res, ret.a[i][0]);
  cout << res << endl;
  return 0;
}
int main() { return Main(); }