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 maxn = 32;
const int maxb = 60;
const int mod = 1e9 + 7;
const long long inf = 2e18;
inline void mkay(int& a) {
  if (a >= mod)
    a -= mod;
  else if (a < 0)
    a += mod;
}
struct matrix {
  int val[maxn][maxn];
  int n;
  matrix(int nn = 0, int x = 0) {
    n = nn;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        if (i != j)
          val[i][j] = 0;
        else
          val[i][j] = x;
  }
  int* operator[](int x) { return val[x]; }
  matrix operator*(matrix b) {
    matrix ans(n, 0);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        for (int k = 0; k < n; k++)
          mkay(ans[i][k] += 1LL * val[i][j] * b[j][k] % mod);
    return ans;
  }
  matrix operator+(matrix b) {
    matrix ans(n, 0);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) ans[i][j] = (val[i][j] + b[i][j]);
    return ans;
  }
  bool operator==(matrix b) {
    if (n != b.n) return 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++)
        if (val[i][j] != b[i][j]) return 0;
    return 1;
  }
};
matrix operator^(matrix a, long long b) {
  matrix ans(a.n, 1);
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a;
    b >>= 1;
  }
  return ans;
}
int k;
matrix a[maxb], shift[maxb];
long long val[maxb];
matrix get(long long n, int ind) {
  if (!n) return matrix(k + 1, 1);
  for (int i = ind; i >= 0; i--)
    if (n >= val[i]) {
      matrix tmp = get(n % val[i], i - 1);
      int x = n / val[i];
      matrix ans = a[i];
      for (int j = 1; j < x; j++) {
        ans = ans * shift[k - 1];
        ans = ans * a[i];
      }
      ans = ans * shift[k - 1];
      ans = ans * tmp;
      ans = ans * shift[x - 1];
      return ans;
    }
  cout << 1 / 0 << endl;
  return matrix(k + 1, -800);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n >> k;
  val[0] = 1;
  int last = maxb;
  for (int i = 1; i < maxb; i++) {
    if (val[i - 1] >= inf / k) {
      val[i] = inf;
      last = i;
      break;
    } else
      val[i] = val[i - 1] * k;
  }
  shift[0] = matrix(k + 1, 1);
  shift[1] = matrix(k + 1);
  for (int i = 0; i < k; i++) shift[1][i][(i - 1 + k) % k] = 1;
  shift[1][k][k] = 1;
  for (int i = 2; i < k; i++) shift[i] = shift[i - 1] * shift[1];
  for (int i = 0; i < maxb; i++) a[i] = matrix(k + 1, 1);
  for (int i = 0; i <= k; i++) a[0][i][0] = 1;
  for (int i = 1; i < last; i++) {
    matrix tmp = a[i - 1];
    a[i] = a[i - 1];
    for (int j = 1; j < k; j++) {
      a[i] = a[i] * shift[k - 1];
      a[i] = a[i] * tmp;
    }
    a[i] = a[i] * shift[k - 1];
  }
  matrix ans(k + 1);
  ans[0][k] = 1;
  ans = ans * get(n, last - 1);
  int res = 0;
  for (int i = 0; i <= k; i++) mkay(res += ans[0][i]);
  cout << res << endl;
}