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;
#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
const long long maxn = 25 + 10, maxm = 1e3 + 10, lg = 21, mod = 1e9 + 7,
                inf = 1e18;
long long n, m;
struct MAT {
  long long a[maxn][maxn];
  MAT() { memset(a, 0, sizeof(a)); }
};
MAT I() {
  MAT r;
  for (int i = 0; i <= m; i++) r.a[i][i] = 1;
  return r;
}
MAT operator*(const MAT& a, const MAT& b) {
  MAT r;
  for (int i = 0; i <= m; i++)
    for (int j = 0; j <= m; j++)
      for (int k = 0; k <= m; k++)
        r.a[i][j] = (r.a[i][j] + a.a[i][k] * b.a[k][j]) % mod;
  return r;
}
MAT dp[maxn << 1][maxn], g1[maxn], g2[maxn];
long long a[maxn << 1];
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  long long nn = 0;
  while (n) a[nn++] = n % m, n /= m;
  nn--;
  for (int i = 0; i < m; i++) {
    dp[0][i] = I();
    for (int j = 0; j <= m; j++) dp[0][i].a[j][i] = 1;
  }
  for (int i = 1; i <= nn; i++) {
    g1[0] = g2[m] = I();
    for (int j = 1; j < m; j++) g1[j] = g1[j - 1] * dp[i - 1][j - 1];
    for (int j = m - 1; j >= 0; j--) g2[j] = dp[i - 1][j] * g2[j + 1];
    for (int j = 0; j < m; j++) dp[i][j] = g2[j] * g1[j];
  }
  MAT res;
  res.a[0][m] = 1;
  for (int i = nn, nw = 0; i >= 0; i--)
    for (int j = 0; j < a[i]; j++) res = res * dp[i][nw], nw = (nw + 1) % m;
  long long ans = 0;
  for (int i = 0; i <= m; i++) ans = (ans + res.a[0][i]) % mod;
  cout << ans;
  return 0;
}