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 = 1000000007;
struct node {
  long long t[35][35];
  long long tot[35];
  int shift;
};
long long N;
int K;
node merge(node L, node R, int shift) {
  node ans = {};
  ans.shift = L.shift;
  for (int i = 0; i < K; i++) {
    for (int j = 0; j < K; j++) {
      long long tot[35] = {0};
      int en = (j - shift - L.shift + K * 2) % K;
      tot[0] = R.t[(K - shift) % K][en];
      for (int k = 1; k < K; k++) {
        tot[k] = (tot[k - 1] + R.t[(K + k - shift) % K][en]) % MOD;
      }
      for (int k = 0; k < K; k++) {
        ans.t[i][j] =
            (ans.t[i][j] + L.t[i][k] * tot[(k - L.shift + K) % K]) % MOD;
      }
    }
  }
  for (int i = 0; i < K; i++) {
    ans.tot[i] = L.tot[i];
    for (int k = 0; k < K; k++) {
      for (int l = 0; l < K; l++) {
        if ((k - L.shift + K) % K < (l + shift) % K) continue;
        ans.tot[i] = (ans.tot[i] + L.t[i][k] * R.tot[l]) % MOD;
      }
    }
  }
  return ans;
}
int main() {
  scanf("%lld%d", &N, &K);
  node mul = {}, ans = {};
  mul.shift = 0;
  for (int i = 0; i < K; i++) {
    mul.tot[i] = 1 << K - i - 1;
    for (int j = i; j < K; j++) {
      if (i == j)
        mul.t[i][j] = 1;
      else
        mul.t[i][j] = 1 << j - i - 1;
    }
  }
  for (int i = 0; i < N % K; i++) {
    ans.tot[i] = 1 << (N % K - i - 1);
    for (int j = i; j < N % K; j++) {
      if (i == j)
        ans.t[i][j] = 1;
      else
        ans.t[i][j] = 1 << j - i - 1;
    }
  }
  int nw = 1;
  N /= K;
  while (N) {
    node tmp = mul;
    for (int i = 0; i < N % K; i++) ans = merge(tmp, ans, nw % K);
    for (int i = 1; i < K; i++) {
      mul = merge(mul, tmp, nw % K);
      mul.shift = (mul.shift + tmp.shift + nw) % K;
    }
    N /= K;
    nw++;
  }
  long long tot = 1;
  for (int i = 0; i < K; i++) tot += ans.tot[i];
  printf("%lld\n", tot % MOD);
  return 0;
}