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 = 1000000007;
vector<vector<vector<int> > > dp;
vector<vector<int> > tmp;
vector<vector<int> > base;
int get_dig(int pos, int k) {
  return (pos < k) ? pos : (get_dig(pos / k, k) + pos % k) % k;
}
inline int get_val(const vector<vector<int> > &a, int i, int j, int shift,
                   int k) {
  if (i == k && j < k) {
    return a[k][(j + k - shift) % k];
  }
  return a[(i + k - shift) % k][(j + k - shift) % k];
}
vector<vector<int> > shift(const vector<vector<int> > &a, int shift) {
  int k = int(a.size()) - 1;
  vector<vector<int> > res = a;
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j < k; ++j) {
      res[i][j] = get_val(a, i, j, shift, k);
    }
  }
  return res;
}
vector<vector<int> > operator*(const vector<vector<int> > &a,
                               const vector<vector<int> > &b) {
  int n = int(a.size()), k = int(a[0].size()), m = int(b[0].size());
  vector<vector<int> > res(n, vector<int>(m, 0));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < m; ++j) {
      for (int l = 0; l < k; ++l) {
        res[i][j] = (res[i][j] + 1ll * a[i][l] * b[l][j]) % MOD;
      }
    }
  }
  return res;
}
int main() {
  long long n;
  int k;
  cin >> n >> k;
  int max_depth = 0;
  {
    long long tmp = n;
    while (tmp != 0) {
      ++max_depth;
      tmp /= k;
    }
  }
  dp.resize(max_depth);
  {
    dp[0].assign(k + 1, vector<int>(k + 1, 0));
    for (int i = 0; i <= k; ++i) {
      for (int j = 0; j <= k; ++j) {
        dp[0][i][j] = (i == j || j == 0);
      }
    }
    vector<vector<int> > e(k + 1, vector<int>(k + 1, 0));
    for (int i = 0; i <= k; ++i) {
      e[i][i] = 1;
    }
    for (int depth = 1; depth < max_depth; ++depth) {
      dp[depth] = e;
      vector<vector<int> > bs = dp[depth - 1];
      vector<vector<int> > &res = dp[depth];
      int curx = k;
      for (int i = 0; (1 << i) <= curx; ++i) {
        if (curx & (1 << i)) {
          res = bs * shift(res, (1 << i));
          curx ^= (1 << i);
        }
        bs = bs * shift(bs, (1 << i));
      }
    }
  }
  base.assign(1, vector<int>(k + 1, 0));
  base[0][k] = 1;
  int sl = 0;
  long long x = n;
  vector<int> vec;
  while (x != 0) {
    vec.push_back(x % k);
    x /= k;
  }
  int len = int(vec.size());
  for (int cur_depth = len - 1; cur_depth >= 0; --cur_depth) {
    int rest = vec[cur_depth];
    for (int it = 0; it < rest; ++it, sl = (sl + 1) % k) {
      base = base * shift(dp[cur_depth], sl);
    }
  }
  {
    int res = 0;
    for (int i = 0; i <= k; ++i) {
      res = (res + base[0][i]) % MOD;
    }
    cout << res << '\n';
  }
  return 0;
}