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 maxn = 35, P = 1000000007;
long long n;
unsigned long long pw[100];
int K;
struct mat {
  int a[maxn][maxn];
  mat() {
    for (int i = 0; i <= K; i++) {
      for (int j = 0; j <= K; j++) a[i][j] = 0;
    }
  }
  mat operator*(const mat &o) const {
    mat ans;
    for (int i = 0; i <= K; i++) {
      for (int j = 0; j <= K; j++)
        if (a[i][j]) {
          for (int k = 0; k <= K; k++) {
            ans.a[i][k] = (ans.a[i][k] + 1LL * a[i][j] * o.a[j][k]) % P;
          }
        }
    }
    return ans;
  }
} f[100][maxn];
int main() {
  scanf("%lld %d", &n, &K);
  for (int i = 0; i < K; i++) {
    for (int j = 0; j <= K; j++)
      if (i ^ j) f[0][i].a[j][j] = 1;
    f[0][i].a[K][i]++;
    f[0][i].a[i][K] = P - 1, f[0][i].a[K][K]++;
  }
  int num = 0;
  long long m = n;
  while (m) num++, m /= K;
  for (int i = pw[0] = 1; i <= num; i++) {
    pw[i] = pw[i - 1] * K;
    f[i][0] = f[i - 1][0];
    for (int j = 1; j < K; j++) {
      f[i][0] = f[i][0] * f[i - 1][j];
    }
    for (int j = 1; j < K; j++) {
      for (int a = 0; a <= K; a++) {
        for (int b = 0; b <= K; b++) {
          int c = a < K ? (a + j) % K : a;
          int d = b < K ? (b + j) % K : b;
          f[i][j].a[c][d] = f[i][0].a[a][b];
        }
      }
    }
  }
  mat A;
  A.a[0][K] = 1;
  int cur = 0;
  for (int i = num; ~i; i--)
    while (n >= pw[i]) {
      A = A * f[i][++cur %= K], n -= pw[i];
    }
  printf("%d\n", A.a[0][K]);
  return 0;
}