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 = 1e9 + 7, N = 33;
long long n;
int k;
vector<int> dig;
inline void upd(long long &a, long long b) {
  a += b;
  if (a >= MOD) a -= MOD;
}
struct M {
  long long a[N][N];
  M() { memset(a, 0, sizeof(a)); }
  void Set() {
    for (int i = (0); i < (k + 1); i++) a[i][i] = 1;
  }
  M operator*(const M &c) const {
    M res;
    for (int i = (0); i < (k + 1); i++)
      for (int j = (0); j < (k + 1); j++)
        for (int _ = (0); _ < (k + 1); _++)
          upd(res.a[i][j], a[i][_] * c.a[_][j] % MOD);
    return res;
  }
} A[N << 1][N], S[N << 1][N], P[N << 1][N], ans;
int main() {
  scanf("%lld%d", &n, &k);
  for (int i = (0); i < (k); i++) {
    A[0][i].Set();
    for (int j = (0); j < (k + 1); j++) A[0][i].a[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; --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) {
    dig.push_back(n % k);
    n /= k;
  }
  int s = 0;
  ans.Set();
  for (int i = (int)dig.size() - 1; ~i; --i) {
    for (int j = (0); j < (dig[i]); j++) {
      ans = ans * A[i][(j + s) % k];
    }
    s = (s + dig[i]) % k;
  }
  long long res = 1;
  for (int i = (0); i < (k); i++) upd(res, ans.a[i][k]);
  printf("%lld\n", res);
  return 0;
}