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 = 31;
const int mod = 1000000007;
struct ma {
  long long a[maxn][maxn];
  int n;
  ma(int m = 0) {
    n = m;
    memset(a, 0, sizeof(a));
    for (int i = 0; i < n; i++) a[i][i] = 1;
  }
  ma(const ma& nn) {
    n = nn.n;
    memcpy(a, nn.a, sizeof(a));
  }
  ma operator*(const ma& nn) {
    ma ret(n);
    memset(ret.a, 0, sizeof(ret.a));
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) {
        for (int k = 0; k < n; k++) ret.a[i][j] += a[i][k] * nn.a[k][j] % mod;
        ret.a[i][j] %= mod;
      }
    return ret;
  }
};
ma mypow(ma a, long long n) {
  ma ret(a.n);
  while (n) {
    if (n & 1) ret = ret * a;
    n >>= 1;
    a = a * a;
  }
  return ret;
}
ma m1[100], m2[100];
int main() {
  long long n;
  int k, i, j;
  cin >> n >> k;
  n--;
  m1[1].n = k + 1;
  for (i = 0; i <= k; i++) m1[1].a[i][i] = 1;
  for (i = 1; i < 99; i++) {
    int tmp = ((-i - 1) % k + k) % k + 1;
    tmp = k - tmp;
    memcpy(m2[i].a[0], m1[i].a[k], sizeof(m1[i].a[k]));
    for (j = 1; j <= k; j++) {
      for (int c = 0; c <= k; c++)
        m2[i].a[j][c] =
            ((m2[i].a[j - 1][c] * 2 - m1[i].a[tmp][c]) % mod + mod) % mod;
      tmp++;
      if (tmp >= k) tmp = 0;
    }
    m2[i].n = k + 1;
    m1[i + 1] = m1[i] * mypow(m2[i], k - 1);
  }
  ma ans(k + 1);
  int ind = 1, sheng = n % k;
  n /= k;
  while (n) {
    ans = ans * mypow(m2[ind], n % k);
    ind++, n /= k;
  }
  long long ret = 0;
  for (i = 0; i <= k; i++) ret = (ret + ans.a[sheng + 1][i] * (1ll << i)) % mod;
  cout << ret << endl;
}