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;
long long n;
int m, K, sum, now, a[110];
void ADD(int &x, int y, int m) {
  x += y;
  if (x >= m) x -= m;
}
struct Matrix {
  int v[35][35];
  Matrix() { memset(v, 0, sizeof(v)); }
  Matrix operator*(const Matrix &a) const {
    Matrix res;
    for (int k = 0; k <= K; k++)
      for (int i = 0; i <= K; i++)
        if (v[i][k])
          for (int j = 0; j <= K; j++)
            if (a.v[k][j])
              res.v[i][j] = (1ll * v[i][k] * a.v[k][j] + res.v[i][j]) % Mod;
    return res;
  }
} ans, s[75][35], pre[75][35], suf[75][35];
void work(long long n) {
  long long tmp = n;
  while (tmp) {
    a[++m] = tmp % K;
    tmp /= K;
  }
  reverse(a + 1, a + m + 1);
}
int main() {
  scanf("%lld%d", &n, &K);
  work(n);
  for (int i = 0; i < K; i++) {
    for (int j = 0; j <= K; j++) s[0][i].v[j][j] = 1;
    for (int j = 0; j <= K; j++) s[0][i].v[i][j] = 1;
  }
  for (int i = 1; i <= m; i++) {
    pre[i - 1][0] = s[i - 1][0];
    suf[i - 1][K - 1] = s[i - 1][K - 1];
    for (int j = 1; j < K; j++) pre[i - 1][j] = pre[i - 1][j - 1] * s[i - 1][j];
    for (int j = K - 2; j >= 0; j--)
      suf[i - 1][j] = s[i - 1][j] * suf[i - 1][j + 1];
    for (int j = 0; j < K; j++)
      if (j)
        s[i][j] = suf[i - 1][j] * pre[i - 1][j - 1];
      else
        s[i][j] = suf[i - 1][j];
  }
  for (int i = 0; i <= K; i++)
    for (int j = 0; j <= K; j++) ans.v[i][j] = (i == j);
  now = 0;
  for (int i = 1; i <= m; i++) {
    if (a[i]) {
      if (now + a[i] - 1 >= K) {
        for (int j = now; j < K; j++) ans = ans * s[m - i][j];
        for (int j = 0; j <= (now + a[i] - 1) - K; j++) ans = ans * s[m - i][j];
      } else
        for (int j = now; j <= now + a[i] - 1; j++) ans = ans * s[m - i][j];
      ADD(now, a[i], K);
    }
  }
  for (int i = 0; i <= K; i++) ADD(sum, ans.v[i][K], Mod);
  printf("%d\n", sum);
  return 0;
}