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 T = 31;
const int mod = 1000000007;
const long long oo = 1LL << 60;
long long N;
int K;
long long B[T][T];
long long power[70];
long long dp[70][T][T];
bool bio[70];
long long rot[T][T][T];
long long mult(long long a, long long b) {
  if (a == 0) return 0;
  if (b >= oo / a + 1) return oo;
  return a * b;
}
void mult(long long A[T][T], long long B[T][T]) {
  static long long C[T][T];
  for (int i = (0); i < (T); ++i)
    for (int j = (0); j < (T); ++j) C[i][j] = 0;
  for (int i = (0); i < (T); ++i)
    for (int j = (0); j < (T); ++j)
      for (int k = (0); k < (T); ++k)
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;
  for (int i = (0); i < (T); ++i)
    for (int j = (0); j < (T); ++j) A[i][j] = C[i][j];
}
void rec(long long prefix, int n, long long A[T][T]) {
  long long lo = mult(prefix, power[n]);
  long long hi = lo + power[n] - 1;
  if (lo >= N) return;
  long long z = prefix % K;
  mult(A, rot[z]);
  if (n == 0) {
    mult(A, B);
  } else {
    bool memo = hi < N;
    auto AA = memo ? dp[n] : A;
    if (!bio[n] && memo)
      for (int i = (0); i < (T); ++i) AA[i][i] = 1;
    if (!bio[n] || !memo)
      for (int i = (0); i < (K); ++i) rec(prefix * K + i, n - 1, AA);
    if (memo) bio[n] = true, mult(A, AA);
  }
  mult(A, rot[(K - z) % K]);
}
int main() {
  scanf("%lld%d", &N, &K);
  for (int i = (0); i < (K); ++i) B[i][i] = 1;
  for (int i = (1); i < (K); ++i) B[i][0] = 1;
  B[K][0] = 1;
  B[K][K] = 1;
  for (int x = (0); x < (K); ++x)
    for (int i = (0); i < (K); ++i) rot[x][i][(i + x) % K] = 1;
  for (int x = (0); x < (K); ++x)
    for (int i = (K); i < (T); ++i) rot[x][i][i] = 1;
  static long long A[T][T];
  A[0][K] = 1;
  power[0] = 1;
  for (int i = (1); i < (70); ++i) power[i] = mult(power[i - 1], K);
  rec(0, 69, A);
  long long ans = 1;
  for (int i = (0); i < (K); ++i) ans = (ans + A[0][i]) % mod;
  printf("%lld\n", ans);
  return 0;
}