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 mo = 1000000007;
long long n, ans[40], Ans, nn;
int K, h2[40];
struct O {
  long long a[40][40];
} E, A[40], B[70], C;
O operator*(const O &A, const O &B) {
  for (int i = 0; i <= K; ++i)
    for (int j = 0; j <= K; ++j) {
      C.a[i][j] = 0;
      for (int k = 0; k <= K; ++k) (C.a[i][j] += A.a[i][k] * B.a[k][j]) %= mo;
    }
  return C;
}
int main() {
  scanf("%lld%d", &n, &K);
  if (n <= K) {
    printf("%lld\n", (1ll << n) % mo);
    return 0;
  }
  for (int t = 0; t < K; ++t) {
    for (int i = 0; i < K; ++i) h2[i] = (i + t) % K;
    for (int j = 0; j < K; ++j) {
      for (int k = 0; k < j; ++k)
        for (int i = 0; i <= K; ++i) (A[t].a[i][j] += A[t].a[i][k]) %= mo;
      for (int i = K - 1; ~i; --i) {
        ++A[t].a[i][j];
        if (i == h2[j]) break;
      }
    }
    for (int j = 0; j <= K; ++j) A[t].a[j][K] = 1;
  }
  nn = n / K;
  B[0] = A[1];
  for (int j = 0; j <= K; ++j) E.a[j][j] = 1;
  for (int j = 1; j < K; ++j) E = B[0] * E;
  for (long long NN = nn / K, i = 1; NN; NN /= K, ++i) {
    B[i] = E * A[(i + 1) % K];
    for (int j = 1; j < K; ++j) E = B[i] * E;
  }
  memset(E.a, 0, sizeof E.a);
  for (int j = 0; j <= K; ++j) E.a[j][j] = 1;
  for (int i = 0; nn; nn /= K, ++i)
    for (int j = nn % K; j; --j) E = B[i] * E;
  for (int i = 0; i < K; ++i)
    for (int j = 0; j <= K; ++j) (ans[j] += (1ll << i) * E.a[i][j]) %= mo;
  Ans = 1 + ans[K];
  for (int j = 0; j < n % K; ++j) (Ans += ans[j]) %= mo;
  printf("%lld\n", (Ans % mo + mo) % mo);
  return 0;
}