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;
long long power(long long a, long long b, long long md) {
  return (!b ? 1
             : (b & 1 ? a * power(a * a % md, b / 2, md) % md
                      : power(a * a % md, b / 2, md) % md));
}
const int xn = 30 + 10;
const int xm = -20 + 10;
const int sq = 320;
const int inf = 1e9 + 10;
const long long INF = 1e18 + 10;
const long double eps = 1e-15;
const int mod = 1e9 + 7;
const int base = 257;
struct matrix {
  int n, m, a[xn][xn];
  matrix operator*(const matrix &t) {
    matrix r;
    r.n = n, r.m = t.m;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < t.m; ++j) {
        r.a[i][j] = 0;
        for (int k = 0; k < m; ++k)
          r.a[i][j] = (r.a[i][j] + 1ll * a[i][k] * t.a[k][j] % mod) % mod;
      }
    }
    return r;
  }
};
long long n;
int k, m, sum, ans;
vector<int> vec;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cin >> n >> k;
  while (n) vec.push_back(n % k), n /= k, ++m;
  matrix T[m];
  for (int i = 0; i < m; ++i) T[i].n = T[i].m = k + 1;
  for (int i = 0; i < m; ++i)
    for (int j = 0; j < xn; ++j)
      for (int l = 0; l < xn; ++l) T[i].a[j][l] = 0;
  for (int i = 0; i <= k; ++i) T[0].a[i][i] = 1;
  for (int i = 0; i <= k; ++i) T[0].a[i][0] = 1;
  matrix S[k];
  for (int i = 0; i < k; ++i) {
    S[i].n = S[i].m = k + 1;
    for (int j = 0; j < k; ++j) S[i].a[j][(i + j) % k] = 1;
    S[i].a[k][k] = 1;
  }
  for (int i = 1; i < m; ++i) {
    for (int j = 0; j <= k; ++j) T[i].a[j][j] = 1;
    for (int j = 0; j < k; ++j) T[i] = (T[i] * T[i - 1]) * S[1];
  }
  matrix M;
  for (int j = 0; j < xn; ++j)
    for (int l = 0; l < xn; ++l) M.a[j][l] = 0;
  M.n = 1, M.m = k + 1;
  M.a[0][k] = 1;
  for (int i = m - 1; i >= 0; --i) {
    for (int j = 0; j < vec[i]; ++j)
      M = ((M * S[(sum + j) % k]) * T[i]) * S[(k + k - sum - j) % k];
    sum = (sum + vec[i]) % k;
  }
  for (int i = 0; i <= k; ++i) ans = (ans + M.a[0][i]) % mod;
  cout << ans << '\n';
  return 0;
}