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>
const int P = 1e9 + 7;
long long ri() {
  char c = getchar();
  long long x = 0, f = 1;
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') f = -1;
  for (; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) - '0' + c;
  return x * f;
}
int k;
struct Maxtir {
  int m[31][31];
  int *operator[](int x) { return m[x]; }
  void I() {
    for (int i = 0; i <= k; ++i) m[i][i] = 1;
  }
  Maxtir operator*(const Maxtir b) {
    static Maxtir c;
    for (int i = 0; i <= k; ++i)
      for (int j = 0; j <= k; ++j) {
        int res = 0;
        for (int l = 0; l <= k; ++l)
          res = (res + 1LL * m[i][l] * b.m[l][j]) % P;
        c[i][j] = res;
      }
    return c;
  }
  void print() {
    for (int i = 0; i <= k; ++i, puts(""))
      for (int j = 0; j <= k; ++j) printf("%d ", m[i][j]);
    puts("");
  }
} A[101][30], suf[30], pre[30], ans;
int d[101], tp;
int main() {
  long long n = ri();
  k = ri();
  for (int i = 0; i < k; ++i) {
    A[0][i].I();
    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) {
    suf[k - 1] = A[i - 1][k - 1];
    for (int j = k - 2; ~j; --j) suf[j] = A[i - 1][j] * suf[j + 1];
    pre[0] = A[i - 1][0];
    for (int j = 1; j < k; ++j) pre[j] = pre[j - 1] * A[i - 1][j];
    A[i][0] = suf[0];
    for (int j = 1; j < k; ++j) A[i][j] = suf[j] * pre[j - 1];
  }
  for (; n; n /= k) d[tp++] = n % k;
  int nw = 0;
  ans.I();
  for (int i = tp - 1; ~i; --i) {
    for (int j = 0; j < d[i]; ++j) ans = ans * A[i][(nw + j) % k];
    nw = (nw + d[i]) % k;
  }
  long long res = 1;
  for (int i = 0; i < k; ++i) res += ans[i][k];
  printf("%d\n", res % P);
  return 0;
}