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>
#pragma GCC optimize("unroll-loops")
using namespace std;
const int MXN = 3e5 + 10;
const int MXZ = 32;
const int Mod = 1e9 + 7;
void mkay(int &x) {
  if (x >= Mod) x -= Mod;
}
long long n;
int k, m, s, ans;
struct Matrix {
  int n, m;
  int M[MXZ][MXZ];
  Matrix(int _n = 0, int _m = 0, int f = 0) {
    n = _n, m = _m;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) {
        M[i][j] = (f == -1 ? (i == j) : f);
      }
  }
  Matrix operator*(const Matrix &T) {
    assert(m == T.n);
    Matrix R(n, T.m);
    for (int i = 0; i < n; i++) {
      for (int k = 0; k < m; k++) {
        for (int j = 0; j < T.m; j++) {
          mkay(R.M[i][j] += 1ll * M[i][k] * T.M[k][j] % Mod);
        }
      }
    }
    return R;
  }
  Matrix Shift(int d = 1) {
    if (d < 0) d = (d + k) % k;
    Matrix R(k + 1, k + 1);
    R.M[k][k] = 1;
    for (int i = 0; i < k; i++) {
      int ip = i + d;
      if (ip >= k) ip -= k;
      for (int j = 0; j < k; j++) {
        int jp = j + d;
        if (jp >= k) jp -= k;
        R.M[ip][jp] = M[i][j];
      }
      R.M[ip][k] = M[i][k];
    }
    return R;
  }
} Ans, C, Cp;
vector<int> digit;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  Matrix Ans(k + 1, 1), C(k + 1, k + 1, -1), Cr(k + 1, k + 1, -1);
  Ans.M[k][0] = 1;
  for (int i = 0; i <= k; i++) C.M[0][i] = 1;
  while (n) {
    int d = n % k;
    s += d;
    digit.push_back(d);
    n /= k, m++;
  }
  s %= k;
  for (int i = 0; i < m; i++) {
    s -= digit[i];
    if (s < 0) s += k;
    for (int j = digit[i] - 1; ~j; --j) {
      Ans = C.Shift((j + s) % k) * Ans;
    }
    Cp = Matrix(k + 1, k + 1, -1);
    for (int j = 0; j < k; j++) Cp = Cp * C.Shift(j);
    swap(C, Cp);
  }
  for (int i = 0; i < k; i++) mkay(ans += Ans.M[i][0]);
  mkay(ans += 1);
  cout << ans << '\n';
  return 0;
}