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;
int nr, K, mod, ans[33], v[70];
long long N;
struct matrix {
  int a[33][33];
  matrix() { memset(a, 0, sizeof(a)); }
  matrix(bool type) {
    memset(a, 0, sizeof(a));
    if (type == 1)
      for (int i = 0; i <= K; i++) a[i][i] = 1;
  }
} dp[70][33], pref[33], suf[33];
matrix operator*(matrix a, matrix b) {
  matrix c;
  for (int i = 0; i <= K; i++)
    for (int j = 0; j <= K; j++) {
      c.a[i][j] = 0;
      for (int k = 0; k <= K; k++)
        c.a[i][j] = ((long long)c.a[i][j] + 1LL * a.a[i][k] * b.a[k][j]) % mod;
    }
  return c;
}
void Prod(matrix a) {
  int old[33];
  for (int i = 0; i <= K; i++) old[i] = ans[i];
  for (int i = 0; i <= K; i++) {
    ans[i] = 0;
    for (int j = 0; j <= K; j++)
      ans[i] = ((long long)ans[i] + 1LL * old[j] * a.a[j][i]) % mod;
  }
}
void Finish() {
  int sum = 0;
  for (int i = 0; i <= K; i++) sum = ((long long)sum + ans[i]) % mod;
  printf("%d\n", sum);
}
int main() {
  scanf("%I64d %d", &N, &K), mod = 1e9 + 7;
  long long aux = N;
  while (aux) v[nr++] = aux % K, aux /= K;
  for (int i = 0; i < K; i++) {
    for (int j = 0; j <= K; j++)
      if (j != i) dp[0][i].a[j][j] = 1;
    for (int j = 0; j <= K; j++) dp[0][i].a[j][i] = 1;
  }
  for (int i = 1; i < nr; i++) {
    pref[0] = dp[i - 1][0];
    for (int j = 1; j < K; j++) pref[j] = pref[j - 1] * dp[i - 1][j];
    suf[K] = matrix(1);
    for (int j = K - 1; j >= 0; j--) suf[j] = dp[i - 1][j] * suf[j + 1];
    dp[i][0] = suf[0];
    for (int j = 1; j < K; j++) dp[i][j] = suf[j] * pref[j - 1];
  }
  ans[K] = 1;
  int curr = 0;
  for (int i = nr - 1; i >= 0; i--)
    while (v[i]--) Prod(dp[i][curr]), curr = (curr + 1) % K;
  Finish();
  return 0;
}