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;
long long N, pw, ans;
long long K, T;
long long sc[35];
struct Matrix {
  long long n, k[35][35];
  Matrix Trans(long long x) {
    long long i, j;
    Matrix c;
    c.n = n;
    for (i = 0; i < K; i++) sc[i] = (i + x) % K;
    sc[K] = K;
    for (i = 0; i < n; i++)
      for (j = 0; j <= K; j++) {
        long long tx = sc[i], ty = sc[j];
        c.k[tx][ty] = k[i][j];
      }
    return c;
  }
  Matrix operator*(Matrix b) const {
    long long i, j, t;
    Matrix c;
    c.n = n;
    for (i = 0; i < n; i++)
      for (j = 0; j <= K; j++)
        for (c.k[i][j] = 0, t = 0; t <= K; t++)
          c.k[i][j] = (c.k[i][j] + (long long)k[i][t] * b.k[t][j]) % 1000000007;
    return c;
  }
} p[65][35], st, ts, ed;
int main() {
  long long i, j, tmp, now;
  scanf("%I64d %I64d", &N, &K);
  st.n = 1, st.k[0][K] = 1;
  p[0][0].n = K + 1;
  for (i = 0; i < K; i++) p[0][0].k[i][0] = 1;
  for (i = 1; i < K; i++) p[0][0].k[i][i] = 1;
  p[0][0].k[K][0] = p[0][0].k[K][K] = 1;
  for (T = 0, pw = 1;; pw *= K, T++) {
    for (i = 1; i < K; i++) p[T][i] = p[T][i - 1] * (p[T][0].Trans(i));
    p[T + 1][0] = p[T][K - 1];
    if (N / K < pw) break;
  }
  for (T, now = 0; T >= 0; pw /= K, T--) {
    tmp = N / pw;
    if (tmp) st = st * p[T][tmp - 1].Trans(now);
    N %= pw;
    now = (now + tmp) % K;
  }
  for (i = 0; i <= K; i++) ans = (ans + st.k[0][i]) % 1000000007;
  printf("%I64d\n", ans);
  return 0;
}