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;
struct matri {
  int m[33][33];
} em, a[70], r, pre[70], suf[70];
long long N, K;
void iprint(matri x) {
  int i, j;
  for (i = 0; i <= K; i++) {
    for (j = 0; j <= K; j++) cout << x.m[i][j] << ' ';
    cout << endl;
  }
  cout << endl;
}
matri imulti(matri a, matri b) {
  int i, j, k;
  matri c = em;
  for (k = 0; k <= K; k++) {
    for (i = 0; i <= K; i++) {
      for (j = 0; j <= K; j++) {
        long long tmp = ((long long)a.m[i][k] * b.m[k][j]) % mod;
        c.m[i][j] = (c.m[i][j] + tmp) % mod;
      }
    }
  }
  return c;
}
void ical() {
  int i;
  pre[0] = a[0];
  for (i = 1; i < K; i++) pre[i] = imulti(pre[i - 1], a[i]);
  suf[K - 1] = a[K - 1];
  for (i = K - 2; i >= 0; i--) suf[i] = imulti(a[i], suf[i + 1]);
  a[0] = suf[0];
  for (i = 1; i < K; i++) a[i] = imulti(suf[i], pre[i - 1]);
}
void iWork() {
  cin >> N >> K;
  int i, j;
  r = em;
  for (i = 0; i < K; i++) {
    a[i] = em;
    for (j = 0; j <= K; j++) a[i].m[j][j] = 1;
    for (j = 0; j <= K; j++) a[i].m[j][i] = 1;
    r.m[i][i] = 1;
  }
  r.m[K][K] = 1;
  for (; N > 0; N /= K) {
    int tt = N % K;
    if (tt) {
      long long pos = 0, tmp = N;
      while (tmp) pos = (pos + (tmp /= K)) % K;
      for (i = tt - 1; i >= 0; i--) {
        r = imulti(a[(pos + i) % K], r);
      }
    }
    ical();
  }
  int ans = 1;
  for (i = 0; i < K; i++) ans = (ans + r.m[K][i]) % mod;
  cout << ans << endl;
}
int main() {
  iWork();
  return 0;
}