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 Max = 40;
const int LOG = 64;
const int Mod = 1e9 + 7;
int K;
struct DP {
  long long A[Max][Max];
  long long* operator[](int id) { return A[id]; }
  friend DP operator*(DP a, DP b) {
    DP c;
    for (int i = 0; i <= K; i++)
      for (int j = 0; j <= K; j++) c[i][j] = 0;
    for (int i = 0; i <= K; i++)
      for (int j = 0; j <= K; j++)
        for (int k = 0; k <= K; k++)
          c[i][k] = (c[i][k] + a[i][j] * b[j][k]) % Mod;
    return c;
  }
  friend DP operator+(DP a, int x) {
    x = (x + K) % K;
    DP b;
    for (int i = 0; i <= K; i++)
      for (int j = 0; j < K; j++) b[i][j] = a[i][(j + x) % K];
    b[K][K] = a[K][K];
    return b;
  }
};
DP dp[LOG][Max], ans;
long long len;
int F(long long num) {
  int ans = 0;
  while (num) {
    ans += num % K;
    num /= K;
  }
  return ans % K;
}
int main() {
  long long N;
  cin >> N >> K;
  for (int i = 0; i < K; i++) {
    dp[0][1][i][0] = 1;
    dp[0][1][i][i] = 1;
  }
  dp[0][1][K][0] = 1;
  dp[0][1][K][K] = 1;
  for (int j = 2; j <= K; j++)
    dp[0][j] = (dp[0][j - 1] + (-j + 1)) * dp[0][1] + (j - 1);
  for (int lg = 1; lg < LOG; lg++) {
    dp[lg][1] = dp[lg - 1][K];
    for (int j = 2; j <= K; j++)
      dp[lg][j] = (dp[lg][j - 1] + (-j + 1)) * dp[lg][1] + (j - 1);
  }
  ans[0][0] = -1;
  int lg = 0;
  long long pw = 1;
  vector<int> vec;
  while (N) {
    int x = N % K;
    vec.push_back(x);
    lg++, pw *= K, N /= K;
  }
  for (; pw /= K, lg--;) {
    if (ans[0][0] == -1)
      ans = dp[lg][vec[lg]];
    else if (vec[lg]) {
      int y = F(len);
      ans = (ans + (-y)) * dp[lg][vec[lg]] + y;
    }
    len += pw * vec[lg];
  }
  long long RES = 0;
  for (int i = 0; i <= K; i++) RES = (RES + ans[K][i]) % Mod;
  cout << RES << '\n';
}