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;
int K, mod = 1000000007;
struct pp {
  int son[32];
} p[3010];
int root;
struct node {
  int a[32][32];
  friend node operator*(node x, node y) {
    int i, j, k;
    node z;
    memset(z.a, 0, sizeof(z.a));
    for (i = 0; i <= K; i++)
      for (j = 0; j <= K; j++)
        for (k = 0; k <= K; k++)
          z.a[i][j] = (z.a[i][j] + (long long)x.a[i][k] * y.a[k][j]) % mod;
    return z;
  }
} A[3010], T[35], W, I;
long long o[110];
node query(int x, int d, long long k) {
  if (o[d] == k) return A[x];
  int i;
  node h = I;
  for (i = 0; i < K; i++) {
    if (k >= o[d - 1]) {
      k -= o[d - 1];
      h = h * A[p[x].son[i]];
    } else {
      if (k) h = h * query(p[x].son[i], d - 1, k);
      break;
    }
  }
  return h;
}
int main() {
  int i, j, N = 1, x, ans = 0;
  long long n;
  scanf("%I64d%d", &n, &K);
  for (i = 0; i <= K; i++) I.a[i][i] = 1;
  o[0] = 1;
  o[1] = K;
  A[K + 1] = I;
  for (i = 0; i < K; i++) {
    for (j = 0; j <= K; j++) A[i + 1].a[j][j] = 1;
    for (j = 0; j <= K; j++) A[i + 1].a[i][j] = 1;
    p[K + 1].son[i] = i + 1;
    A[K + 1] = A[K + 1] * A[i + 1];
  }
  root = K + 1;
  while (o[N] != -1 && o[N] < n) {
    if (o[N] <= n / K)
      o[++N] = o[N - 1] * K;
    else
      o[++N] = -1;
    T[K] = I;
    for (i = K - 1; i >= 0; i--) T[i] = A[p[root].son[i]] * T[i + 1];
    W = I;
    x = root;
    root += K;
    A[root] = A[x];
    p[root].son[0] = x;
    for (i = 1; i < K; i++) {
      for (j = i; j < K; j++) p[x + i].son[j - i] = p[x].son[j];
      for (j = 0; j < i; j++) p[x + i].son[j + K - i] = p[x].son[j];
      W = W * A[p[x].son[i - 1]];
      A[x + i] = T[i] * W;
      A[root] = A[root] * A[x + i];
      p[root].son[i] = x + i;
    }
  }
  W = query(root, N, n);
  for (i = 0; i <= K; i++) ans = (ans + W.a[i][K]) % mod;
  printf("%d\n", ans);
  return 0;
}