Problem: Let S(n) denote the number that represents the digits of n in sorted order. For example, S(1) = 1, S(5) = 5, S(50394) = 3459, S(353535) = 333555.

Given a number X, compute <image> modulo 109 + 7.

Input

The first line of input will contain the integer X (1 ≤ X ≤ 10700).

Output

Print a single integer, the answer to the question.

Examples

Input

21


Output

195


Input

345342


Output

390548434

Note

The first few values of S are 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 12. The sum of these values is 195. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
void getre() {
  int x = 0;
  printf("%d\n", 1 / x);
}
void gettle() {
  int res = 1;
  while (1) res <<= 1;
  printf("%d\n", res);
}
template <typename T, typename S>
inline bool upmin(T &a, const S &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T, typename S>
inline bool upmax(T &a, const S &b) {
  return a < b ? a = b, 1 : 0;
}
template <typename N, typename PN>
inline N flo(N a, PN b) {
  return a >= 0 ? a / b : -((-a - 1) / b) - 1;
}
template <typename N, typename PN>
inline N cei(N a, PN b) {
  return a > 0 ? (a - 1) / b + 1 : -(-a / b);
}
template <typename N>
N gcd(N a, N b) {
  return b ? gcd(b, a % b) : a;
}
template <typename N>
inline int sgn(N a) {
  return a > 0 ? 1 : (a < 0 ? -1 : 0);
}
inline void gn(long long &x) {
  int sg = 1;
  char c;
  while (((c = getchar()) < '0' || c > '9') && c != '-')
    ;
  c == '-' ? (sg = -1, x = 0) : (x = c - '0');
  while ((c = getchar()) >= '0' && c <= '9') x = x * 10 + c - '0';
  x *= sg;
}
inline void gn(int &x) {
  long long t;
  gn(t);
  x = t;
}
inline void gn(unsigned long long &x) {
  long long t;
  gn(t);
  x = t;
}
inline void gn(double &x) {
  double t;
  scanf("%lf", &t);
  x = t;
}
inline void gn(long double &x) {
  double t;
  scanf("%lf", &t);
  x = t;
}
inline void gs(char *s) { scanf("%s", s); }
inline void gc(char &c) {
  while ((c = getchar()) > 126 || c < 33)
    ;
}
inline void pc(char c) { putchar(c); }
inline long long sqr(long long a) { return a * a; }
inline double sqrf(double a) { return a * a; }
const int inf = 0x3f3f3f3f;
const double pi = 3.14159265358979323846264338327950288L;
const double eps = 1e-6;
const int mo = 1e9 + 7;
int qp(int a, long long b) {
  int n = 1;
  do {
    if (b & 1) n = 1ll * n * a % mo;
    a = 1ll * a * a % mo;
  } while (b >>= 1);
  return n;
}
int memo[777][10];
int memo2[777][10];
int fac[4444];
int ifac[4444];
char s[777];
int po[11][777];
int invpo[11][777];
int n;
int main() {
  for (int d = (0), _ed = (11); d < _ed; d++) {
    po[d][0] = 1;
    for (int i = (1), _ed = (777); i < _ed; i++)
      po[d][i] = 1ll * po[d][i - 1] * d % mo;
    for (int i = (0), _ed = (777); i < _ed; i++)
      invpo[d][i] = qp(po[d][i], mo - 2);
  }
  gs(s + 1);
  n = strlen(s + 1);
  fac[0] = 1;
  for (int i = (1), _ed = (4444); i < _ed; i++)
    fac[i] = 1ll * fac[i - 1] * i % mo;
  for (int i = (0), _ed = (4444); i < _ed; i++) ifac[i] = qp(fac[i], mo - 2);
  for (int d = (0), _ed = (10); d < _ed; d++)
    for (int a = (0), _ed = (n + 1); a < _ed; a++)
      for (int b = 0; b + a <= n; b++) {
        (((memo[a + b][d]) = ((memo[a + b][d]) +
                              (1ll * ifac[a] * ifac[b] % mo * invpo[d][a + b] %
                               mo * po[9 - d][b] % mo * po[10][b] % mo *
                               (po[10][a]) % mo * invpo[9][1] % mo * d % mo)) %
                             mo) < 0
             ? (memo[a + b][d]) += mo
             : (memo[a + b][d]));
        (((memo2[a + b][d]) = ((memo2[a + b][d]) +
                               (1ll * ifac[a] * ifac[b] % mo * invpo[d][a + b] %
                                mo * po[9 - d][b] % mo * po[10][b] % mo * (-1) %
                                mo * invpo[9][1] % mo * d % mo)) %
                              mo) < 0
             ? (memo2[a + b][d]) += mo
             : (memo2[a + b][d]));
      }
  int tot = 0;
  for (int t = 1; t <= n; t++) {
    int nex = s[t] - '0';
    if (t != n) nex--;
    for (int da = 0; da <= nex; da++) {
      int num[10] = {0};
      for (int j = 1; j <= t - 1; j++) num[s[j] - '0']++;
      num[da]++;
      for (int d = 1; d <= 9; d++) {
        int L = n - t;
        int p = num[d];
        int q = 0;
        for (int tt = d + 1; tt <= 9; tt++) q += num[tt];
        for (int ab = 0; ab <= L; ab++) {
          int temp = (1ll * memo[ab][d] * po[10][p] + memo2[ab][d]) % mo;
          (((temp) = 1ll * (temp) * (po[d][L]) % mo) < 0 ? (temp) += mo
                                                         : (temp));
          (((temp) = 1ll * (temp) * (fac[L]) % mo) < 0 ? (temp) += mo : (temp));
          (((temp) = 1ll * (temp) * (ifac[L - ab]) % mo) < 0 ? (temp) += mo
                                                             : (temp));
          (((temp) = 1ll * (temp) * (po[10][q]) % mo) < 0 ? (temp) += mo
                                                          : (temp));
          (((tot) = ((tot) + (temp)) % mo) < 0 ? (tot) += mo : (tot));
        }
      }
    }
  }
  printf("%d\n", tot);
  return 0;
}