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>
#pragma GCC optimize("O2,unroll-loops")
using namespace std;
const long double eps = 1e-7;
const int inf = 1000000010;
const long long INF = 10000000000000010LL;
const int mod = 1000000007;
const int K = 30, LOG = 61;
struct Matrix {
  long long M[K + 1][K + 1];
  Matrix() { memset(M, 0, sizeof(M)); }
  long long* operator[](int i) { return M[i]; }
};
Matrix operator*(Matrix A, Matrix B) {
  Matrix C;
  for (int k = 0; k <= K; k++)
    for (int i = 0; i <= K; i++)
      for (int j = 0; j <= K; j++)
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;
  return C;
}
long long n, m, k, u, v, x, y, t, a, b, ans;
Matrix T[LOG][K], Ans;
void Build(long long n, int sum) {
  if (!n) return;
  long long lg = 0, tmp = n;
  while (tmp >= k) {
    tmp /= k;
    lg++;
  }
  for (int i = 0; i < tmp; i++) Ans = Ans * T[lg][(sum + i) % k];
  sum = (sum + tmp) % mod;
  while (lg--) tmp *= k;
  Build(n - tmp, sum);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  for (int j = 0; j < k; j++) {
    for (int i = 0; i < k; i++) T[0][j][i][i] = T[0][j][i][j] = 1;
    T[0][j][K][K] = T[0][j][K][j] = 1;
  }
  for (int lg = 1; lg < LOG; lg++) {
    for (int i = 0; i <= K; i++) T[lg][0][i][i] = 1;
    for (int i = 0; i < k; i++) T[lg][0] = T[lg][0] * T[lg - 1][i];
    for (int x = 1; x < k; x++) {
      for (int i = 0; i < k; i++)
        for (int j = 0; j < k; j++)
          T[lg][x][(i + x) % k][(j + x) % k] = T[lg][0][i][j];
      for (int i = 0; i < k; i++) {
        T[lg][x][(i + x) % k][K] = T[lg][0][i][K];
        T[lg][x][K][(i + x) % k] = T[lg][0][K][i];
      }
      T[lg][x][K][K] = T[lg][0][K][K];
    }
  }
  Ans[0][K] = 1;
  Build(n, 0);
  for (int i = 0; i <= K; i++) ans += Ans[0][i];
  ans %= mod;
  cout << ans << '\n';
  return 0;
}