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 n;
int K;
int mat[110][31][31][31];
int ans[31][31];
void multipy(int A[][31], int B[][31], int C[][31]) {
  static int reslut[31][31];
  for (int i = 0; i <= K; i++)
    for (int j = 0; j <= K; j++) {
      reslut[i][j] = 0;
      for (int k = 0; k <= K; k++) {
        reslut[i][j] += A[i][k] * 1ll * B[k][j] % 1000000007;
        reslut[i][j] %= 1000000007;
      }
    }
  for (int i = 0; i < (K + 1); i++)
    for (int j = 0; j < (K + 1); j++) C[i][j] = reslut[i][j];
}
int solve(long long n, int x) {
  int startPos = 0;
  if (n >= K) {
    static int L[31][31][31], R[31][31][31];
    for (int i = 0; i < (K + 1); i++)
      for (int j = 0; j < (K + 1); j++) L[K][i][j] = R[0][i][j] = i == j;
    for (int k = 0; k < (K); k++) {
      multipy(R[k], mat[x][k], R[k + 1]);
      multipy(mat[x][K - k - 1], L[K - k], L[K - k - 1]);
    }
    for (int k = 0; k < (K); k++)
      multipy(L[K - k], R[K - k], mat[x + 1][(K - k) % K]);
    startPos = solve(n / K, x + 1);
  }
  for (int i = 0; i < (n % K); i++)
    multipy(ans, mat[x][(i + startPos) % K], ans);
  startPos += n % K;
  startPos %= K;
  return startPos;
}
int main() {
  cin >> n >> K;
  for (int k = 0; k < (K); k++) {
    for (int i = 0; i < (K + 1); i++)
      for (int j = 0; j < (K + 1); j++) mat[0][k][i][j] = i == j;
    for (int i = 0; i < (K + 1); i++) mat[0][k][i][k] = 1;
  }
  for (int i = 0; i < (K + 1); i++)
    for (int j = 0; j < (K + 1); j++) ans[i][j] = i == j;
  solve(n, 0);
  int reslut = 0;
  for (int i = 0; i < (K + 1); i++) {
    reslut += ans[K][i];
    reslut %= 1000000007;
  }
  printf("%d\n", reslut);
  return 0;
}