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;
long long N;
int K;
struct MAT {
  int m[31][31];
  MAT operator*(const MAT &ot) const {
    MAT ret = {
        0,
    };
    for (int i = 0; i <= K; i++)
      for (int j = 0; j <= K; j++) {
        ret.m[i][j] = 0;
        for (int k = 0; k <= K; k++) {
          ret.m[i][j] = (ret.m[i][j] + (long long)m[i][k] * ot.m[k][j]) % MOD;
        }
      }
    return ret;
  }
} A[65][30], P[65][30], S[65][30], res;
int main() {
  cin >> N >> K;
  vector<int> base_k;
  for (; N; N /= K) base_k.push_back(N % K);
  for (int i = 0; i < K; i++) {
    for (int j = 0; j <= K; j++) A[0][i].m[j][j] = 1;
    for (int j = 0; j <= K; j++) A[0][i].m[i][j] = 1;
  }
  for (int i = 1; i < ((int)(base_k).size()); i++) {
    P[i - 1][0] = A[i - 1][0];
    for (int j = 1; j < K; j++) P[i - 1][j] = P[i - 1][j - 1] * A[i - 1][j];
    S[i - 1][K - 1] = A[i - 1][K - 1];
    for (int j = K - 1; j--;) S[i - 1][j] = A[i - 1][j] * S[i - 1][j + 1];
    for (int j = 0; j < K; j++) {
      A[i][j] = S[i - 1][j];
      if (j) A[i][j] = A[i][j] * P[i - 1][j - 1];
    }
  }
  int x = 0;
  for (int i = 0; i <= K; i++) res.m[i][i] = 1;
  for (int i = ((int)(base_k).size()); i--;) {
    for (int j = 0; j < base_k[i]; j++) {
      res = res * A[i][x];
      x = (x + 1) % K;
    }
  }
  int ans = 0;
  for (int i = 0; i <= K; i++) ans = (ans + res.m[i][K]) % MOD;
  cout << ans << endl;
}