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 long long mod = 1e9 + 7;
int n, num;
long long V;
long long ans[110], dog[101];
vector<int> bit;
struct node {
  long long a[110][110];
  node() { memset(a, 0, sizeof(a)); }
} D, A[88];
node mul(node a, node b) {
  node c;
  for (int i = 1; i <= n + 1; ++i)
    for (int j = 1; j <= n + 1; ++j)
      for (int k = 1; k <= n + 1; ++k)
        (c.a[i][j] += a.a[i][k] * b.a[k][j] % mod) %= mod;
  return c;
}
node move(node a, long long k) {
  node b = a;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      b.a[(k + i - 1) % n + 1][(k + j - 1) % n + 1] = a.a[i][j];
  for (int i = 1; i <= n; ++i) {
    b.a[n + 1][(k + i - 1) % n + 1] = a.a[n + 1][i];
    b.a[(k + i - 1) % n + 1][n + 1] = a.a[i][n + 1];
  }
  return b;
}
int main() {
  scanf("%lld%d", &V, &n);
  for (int i = 1; i <= n + 1; ++i) D.a[i][i] = 1;
  A[0] = D;
  for (int i = 1; i <= n + 1; ++i) A[0].a[1][i] = 1;
  num = 0;
  while (V) {
    bit.push_back(V % n);
    V /= n;
    ++num;
    A[num] = D;
    node B = A[num - 1];
    int k = n, id = 0;
    while (k) {
      if (k % 2 == 1) A[num] = mul(A[num], move(B, (k - 1) << id));
      B = mul(move(B, 1 << id), B);
      k /= 2;
      ++id;
    }
  }
  ans[n + 1] = 1;
  int len = bit.size(), sum = 0;
  for (int I = len - 1; I >= 0; --I) {
    for (int J = 0; J < bit[I]; ++J) {
      node B = move(A[I], sum + J);
      for (int i = 1; i <= n + 1; ++i) dog[i] = ans[i], ans[i] = 0;
      for (int i = 1; i <= n + 1; ++i)
        for (int j = 1; j <= n + 1; ++j)
          (ans[i] += dog[j] * B.a[i][j] % mod) %= mod;
    }
    sum += bit[I];
  }
  long long ANS = 0;
  for (int i = 1; i <= n + 1; ++i) (ANS += ans[i]) %= mod;
  printf("%lld\n", ANS);
  return 0;
}