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 = 1000000007;
struct matrix {
  long long a[35][35];
} a[35], b[35], c, d[35];
long long n;
int K;
inline matrix multi(const matrix &a, const matrix &b) {
  matrix c;
  memset(c.a, 0, sizeof(c.a));
  for (int i = 0; i <= K; i++)
    for (int j = 0; j <= K; j++)
      for (int k = 0; k <= K; k++) (c.a[i][j] += a.a[i][k] * b.a[k][j]) %= MOD;
  return c;
}
inline void merge() {
  b[0] = a[0];
  for (int i = 1; i < K; i++) b[i] = multi(b[i - 1], a[i]);
  d[K - 1] = a[K - 1];
  for (int i = K - 2; i >= 0; i--) d[i] = multi(a[i], d[i + 1]);
  for (int i = 0; i < K; i++) a[i] = i ? multi(d[i], b[i - 1]) : d[i];
}
int main() {
  scanf("%I64d%d", &n, &K);
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < K; j++)
      if (i != j)
        a[i].a[j][j] = 1;
      else
        for (int k = 0; k <= K; k++) a[i].a[k][j] = 1;
    a[i].a[K][K] = 1;
  }
  for (int i = 0; i <= K; i++) c.a[i][i] = 1;
  for (; n; n /= K) {
    int A = n % K;
    if (A) {
      int s = 0, ed;
      long long v;
      for (v = n; v; v /= K) s = (v / K + s) % K;
      ed = (s + A - 1) % K;
      for (;;) {
        c = multi(a[ed], c);
        if (ed == s) break;
        if (--ed < 0) ed += K;
      }
    }
    merge();
  }
  long long res = 1;
  for (int i = 0; i < K; i++) (res += c.a[K][i]) %= MOD;
  printf("%I64d\n", res);
  return 0;
}