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 long long MOD = 1000000007;
const int N = 35;
struct Matrix {
  int size;
  long long a[N][N];
  Matrix operator*(const Matrix &b) const {
    Matrix c;
    c.size = size;
    memset(c.a, 0, sizeof(c.a));
    for (int i = 0; i < size; ++i) {
      for (int j = 0; j < size; ++j) {
        for (int k = 0; k < size; ++k) c.a[i][j] += a[i][k] * b.a[k][j] % MOD;
      }
    }
    for (int i = 0; i < size; ++i)
      for (int j = 0; j < size; ++j) c.a[i][j] %= MOD;
    return c;
  }
};
Matrix dp[60][30];
Matrix pre[30], suf[30];
long long res[35], tmp[35];
int dig[60];
void mul(Matrix &a, long long *b) {
  for (int i = 0; i < a.size; ++i) {
    tmp[i] = 0;
    for (int j = 0; j < a.size; ++j) {
      tmp[i] += a.a[i][j] * b[j] % MOD;
    }
    tmp[i] %= MOD;
  }
  for (int i = 0; i < a.size; ++i) {
    b[i] = tmp[i];
  }
}
int main() {
  long long n;
  int k;
  cin >> n >> k;
  int len = 0;
  while (n != 0) {
    dig[len++] = n % k;
    n /= k;
  }
  for (int i = 0; i < len; ++i)
    for (int j = 0; j < k; ++j) dp[i][j].size = k + 1;
  for (int i = 0; i < k; ++i) {
    memset(dp[0][i].a, 0, sizeof(dp[0][i].a));
    for (int j = 0; j <= k; ++j) dp[0][i].a[j][j] = 1;
    for (int j = 0; j <= k; ++j) dp[0][i].a[i][j] = 1;
  }
  for (int i = 1; i < len; ++i) {
    pre[0] = dp[i - 1][0];
    for (int j = 1; j < k; ++j) pre[j] = dp[i - 1][j] * pre[j - 1];
    suf[k - 1] = dp[i - 1][k - 1];
    for (int j = k - 2; j >= 0; --j) suf[j] = suf[j + 1] * dp[i - 1][j];
    for (int j = 0; j < k; ++j) {
      if (j == 0)
        dp[i][j] = pre[k - 1];
      else {
        dp[i][j] = pre[j - 1] * suf[j];
      }
    }
  }
  res[k] = 1;
  int val = 0;
  for (int i = len - 1; i >= 0; --i) {
    for (int j = 0; j < dig[i]; ++j) {
      mul(dp[i][(val + j) % k], res);
    }
    val += dig[i];
    val %= k;
  }
  int ans = 0;
  for (int i = 0; i <= k; ++i) {
    ans += res[i];
    if (ans >= MOD) ans -= MOD;
  }
  cout << ans << endl;
  return 0;
}