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;
struct atom {
  int x[31][31];
} dp[70], B;
int K, mo = 1e9 + 7, x[70], len, A[70];
long long n;
atom mul(atom k1, atom k2) {
  memset(B.x, 0x00, sizeof B.x);
  for (int i = 0; i < K; i++) {
    B.x[i][K] = k2.x[i][K];
    for (int j = 0; j < K; j++) {
      B.x[i][K] = (B.x[i][K] + 1ll * k2.x[i][j] * k1.x[j][K]) % mo;
      for (int k = 0; k < K; k++)
        B.x[i][k] = (B.x[i][k] + 1ll * k2.x[i][j] * k1.x[j][k]) % mo;
    }
  }
  return B;
}
int main() {
  scanf("%I64d%d", &n, &K);
  len = 1;
  x[1] = (n - 1) % K;
  long long kk = (n - 1) / K;
  while (kk) {
    x[++len] = kk % K;
    kk /= K;
  }
  for (int j = 0; j <= len; j++)
    for (int i = 0; i < K; i++) dp[j].x[i][i] = 1;
  for (int j = 0; j <= K; j++) dp[0].x[0][j] = 1;
  for (int i = 1; i <= len; i++) {
    for (int j = 0; j < K; j++) {
      atom k;
      for (int k1 = 0; k1 < K; k1++) A[k1] = (k1 + j) % K;
      for (int k1 = 0; k1 < K; k1++) {
        k.x[A[k1]][K] = dp[i - 1].x[k1][K];
        for (int k2 = 0; k2 < K; k2++) k.x[A[k1]][A[k2]] = dp[i - 1].x[k1][k2];
      }
      dp[i] = mul(dp[i], k);
    }
  }
  atom ans;
  memset(ans.x, 0x00, sizeof ans.x);
  for (int i = 0; i < K; i++) ans.x[i][i] = 1;
  int pre = 0;
  x[1]++;
  for (int i = len; i; i--) {
    for (int j = 0; j < x[i]; j++) {
      int kk = (pre + j) % K;
      atom k;
      for (int k1 = 0; k1 < K; k1++) A[k1] = (k1 + kk) % K;
      for (int k1 = 0; k1 < K; k1++) {
        k.x[A[k1]][K] = dp[i - 1].x[k1][K];
        for (int k2 = 0; k2 < K; k2++) k.x[A[k1]][A[k2]] = dp[i - 1].x[k1][k2];
      }
      ans = mul(ans, k);
    }
    pre = (pre + x[i]) % K;
  }
  int num = 1;
  for (int i = 0; i < K; i++) num = (num + ans.x[i][K]) % mo;
  cout << num << endl;
  return 0;
}