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 = 1e9 + 7;
int add(int a, int b) {
  a += b;
  a += (a < 0) * mod - (a >= mod) * mod;
  return a;
}
int mul(int a, int b) { return 1LL * a * b % mod; }
struct mat {
  int n, m;
  vector<vector<int>> a;
  mat(int x = 0, int y = 0) {
    n = x, m = y;
    a.resize(n, vector<int>(m, 0));
  }
  void identity() {
    assert(n == m);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) a[i][j] = i == j;
  }
  void clear() {
    for (int i = 0; i < n; i++)
      for (int j = 0; j < m; j++) a[i][j] = 0;
  }
  void resize(int x, int y) {
    n = x, m = y;
    a.resize(n, vector<int>(m, 0));
  }
  mat operator*(const mat &b) {
    assert(m == b.n);
    mat res(n, b.m);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < b.m; j++)
        for (int k = 0; k < m; k++)
          res.a[i][j] = add(res.a[i][j], mul(a[i][k], b.a[k][j]));
    return res;
  }
  void operator*=(const mat &b) { *this = *this * b; }
  mat operator^(long long b) {
    mat res(n, m), add = *this;
    for (res.identity(); b; b >>= 1, add *= add)
      if (b & 1) res *= add;
    return res;
  }
  void operator^=(long long b) { *this = *this ^ b; }
};
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  long long n, k;
  cin >> n >> k;
  mat cnt(1, k + 1);
  cnt.a[0][k] = 1;
  mat A[60][k];
  for (int i = 0; i < k; i++) {
    A[0][i].resize(k + 1, k + 1);
    A[0][i].identity();
    for (int j = 0; j <= k; j++) A[0][i].a[j][i] = 1;
  }
  vector<int> vec;
  while (n) {
    vec.push_back(n % k);
    n /= k;
  }
  int x = 0, sz = (int)(vec.size());
  for (int i = 1; i < sz; i++) {
    mat pre[k];
    for (int i = 0; i < k; i++) pre[i].resize(k + 1, k + 1);
    pre[0] = A[i - 1][0];
    for (int j = 1; j < k; j++) pre[j] = pre[j - 1] * A[i - 1][j];
    mat suf[k];
    for (int i = 0; i < k; i++) suf[i].resize(k + 1, k + 1);
    suf[k - 1] = A[i - 1][k - 1];
    for (int j = k - 2; j >= 0; j--) suf[j] = A[i - 1][j] * suf[j + 1];
    for (int j = 0; j < k; j++) {
      A[i][j] = suf[j];
      if (j) A[i][j] *= pre[j - 1];
    }
  }
  for (int i = sz - 1; i >= 0; i--)
    for (int j = 0; j < vec[i]; j++) {
      cnt *= A[i][x];
      (x += 1) %= k;
    }
  int ans = 0;
  for (int i = 0; i <= k; i++) ans = add(ans, cnt.a[0][i]);
  cout << ans << '\n';
  return 0;
}