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 N = 35, M = 100005, mod = 1e9 + 7;
template <class T>
inline void gmin(T &x, const T &y) {
  if (x > y) x = y;
}
template <class T>
inline void gmax(T &x, const T &y) {
  if (x < y) x = y;
}
inline void ch(int &x, int y) { x = (x + y) % mod; }
inline void read(int &x) {
  x = 0;
  char ch = getchar(), rev = 0;
  while (ch > '9' || ch < '0') rev = (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x = rev ? -x : x;
}
inline int exp(int x, int y) {
  int ans = 1;
  while (y) {
    if (y & 1) ans = (long long)ans * x % mod;
    x = (long long)x * x % mod;
    y >>= 1;
  }
  return ans;
}
struct matrix {
  int a[N][N];
} ans, f[N << 1][N], l[N], r[N];
int k;
long long n;
inline matrix operator*(const matrix &a, const matrix &b) {
  matrix tmp;
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= k; j++) {
      long long temp = 0;
      for (int t = 0; t <= k; t++)
        temp += (long long)a.a[i][t] * b.a[t][j] % mod;
      tmp.a[i][j] = temp % mod;
    }
  return tmp;
}
void print(const matrix &a) {
  for (int i = 0; i <= k; i++, puts(""))
    for (int j = 0; j <= k; j++) printf("%d ", a.a[i][j]);
  puts("");
}
int main() {
  scanf("%lld%d", &n, &k);
  int m = 1;
  long long tmp = 1;
  while (tmp < n) tmp *= k, m++;
  for (int i = 0; i < k; i++) {
    matrix &a = f[0][i];
    for (int j = 0; j <= k; j++) a.a[i][j] = a.a[j][j] = 1;
  }
  for (int j = 1; j < m; j++) {
    l[0] = f[j - 1][0];
    for (int i = 1; i < k; i++) l[i] = f[j - 1][i] * l[i - 1];
    r[k - 1] = f[j - 1][k - 1];
    for (int i = k - 2; ~i; i--) r[i] = r[i + 1] * f[j - 1][i];
    for (int i = 0; i < k; i++) {
      f[j][i] = r[i];
      if (i) f[j][i] = l[i - 1] * f[j][i];
    }
  }
  int s = 0;
  for (int i = 0; i <= k; i++) ans.a[i][i] = 1;
  while (n) {
    long long t = 1, lg = 0;
    for (int i = 1; i < m; i++, t *= k)
      if (t <= n / k)
        lg = i;
      else
        break;
    ans = f[lg][s] * ans;
    s = (s + 1) % k;
    n -= t;
  }
  int res = 0;
  for (int i = 0; i <= k; i++) ch(res, ans.a[i][k]);
  printf("%d\n", res);
  return 0;
}