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 N = 31;
template <typename T>
inline void read(T &AKNOI) {
  T x = 0, flag = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') flag = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  AKNOI = flag * x;
}
namespace ModCalculator {
const int MOD = 1e9 + 7;
const long long MOD2 = 1LL * MOD * MOD;
inline void Inc(int &x, int y) {
  x += y;
  if (x >= MOD) x -= MOD;
}
inline void Dec(int &x, int y) {
  x -= y;
  if (x < 0) x += MOD;
}
inline int Add(int x, int y) {
  Inc(x, y);
  return x;
}
inline int Sub(int x, int y) {
  Dec(x, y);
  return x;
}
inline int Mul(int x, int y) { return 1LL * x * y % MOD; }
}  // namespace ModCalculator
using namespace ModCalculator;
long long n;
int K, d[64], len;
struct Matrix {
  int a[N][N];
  inline void E() {
    for (int i = 0; i <= K; ++i) {
      a[i][i] = 1;
    }
  }
  Matrix operator*(const Matrix &b) const {
    Matrix c;
    for (int i = 0; i <= K; ++i) {
      for (int j = 0; j <= K; ++j) {
        long long tmp = 0;
        for (int k = 0; k <= K; ++k) {
          tmp += 1LL * a[i][k] * b.a[k][j];
          if (tmp > MOD2) tmp -= MOD2;
        }
        c.a[i][j] = tmp % MOD;
      }
    }
    return c;
  }
} F[64][N], pre[N], suf[N], Ans;
void init() {
  read(n);
  read(K);
  for (int i = 0; i < K; ++i) {
    F[0][i].E();
    for (int j = 0; j <= K; ++j) {
      F[0][i].a[j][i] = 1;
    }
  }
  long long v = K;
  for (int i = 1; v <= n; v *= K, ++i) {
    pre[0] = F[i - 1][0];
    for (int j = 1; j < K; ++j) {
      pre[j] = pre[j - 1] * F[i - 1][j];
    }
    suf[K - 1] = F[i - 1][K - 1];
    for (int j = K - 2; j > 0; --j) {
      suf[j] = F[i - 1][j] * suf[j + 1];
    }
    F[i][0] = pre[K - 1];
    for (int j = 1; j < K; ++j) {
      F[i][j] = suf[j] * pre[j - 1];
    }
  }
}
void solve() {
  for (; n; n /= K) {
    d[len++] = n % K;
  }
  Ans.E();
  for (int i = len - 1, x = 0; i >= 0; --i) {
    for (int j = 0; j < d[i]; ++j) {
      Ans = Ans * F[i][(j + x) % K];
    }
    x = (x + d[i]) % K;
  }
  int ans = 1;
  for (int i = 0; i < K; ++i) {
    Inc(ans, Ans.a[K][i]);
  }
  printf("%d\n", ans);
}
int main() {
  init();
  solve();
  return 0;
}