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;
const int N = 710, size = 1 << 20, mod = (int)1e9 + 7, inf = 2e9;
const long long INF = 1e15;
template <class o>
void qr(o& x) {
  char c = getchar();
  x = 0;
  int f = 1;
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) x = x * 10 + c - '0', c = getchar();
  x *= f;
}
template <class o, class... O>
void qr(o& x, O&... y) {
  qr(x);
  qr(y...);
}
template <class o>
void qw(o x) {
  if (x / 10) qw(x / 10);
  putchar(x % 10 + '0');
}
template <class o>
void pr1(o x) {
  if (x < 0) x = -x, putchar('-');
  qw(x);
  putchar(' ');
}
template <class o, class... O>
void pr1(o x, O... y) {
  pr1(x);
  pr1(y...);
}
template <class o>
void pr2(o x) {
  if (x < 0) x = -x, putchar('-');
  qw(x);
  putchar('\n');
}
template <class o, class... O>
void pr2(o x, O... y) {
  pr2(x);
  pr2(y...);
}
long long gcd(long long a, long long b) { return !a ? b : gcd(b % a, a); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long power(long long a, long long b = mod - 2, long long p = mod) {
  long long c = 1;
  while (b) {
    if (b & 1) c = c * a % p;
    b /= 2;
    a = a * a % p;
  }
  return c;
}
long long exgcd(long long a, long long b, long long& x, long long& y) {
  if (!a) {
    x = 0;
    y = 1;
    return b;
  }
  long long d = exgcd(b % a, a, y, x);
  x -= b / a * y;
  return d;
}
long long INV(long long n, long long p = mod) {
  long long x, y;
  exgcd(n, p, x, y);
  return (x % p + p) % p;
}
template <class o>
void cmax(o& x, o y) {
  if (x < y) x = y;
}
template <class o>
void cmin(o& x, o y) {
  if (x > y) x = y;
}
template <typename t1, typename t2>
void ad(t1& x, t2 y) {
  x += y;
  if (x >= mod) x -= mod;
}
template <typename t1, typename t2>
void dl(t1& x, t2 y) {
  x -= y;
  if (x < 0) x += mod;
}
char str[N];
int n, a[N], cnt[10], f[N][N][10][2];
void solve() {
  scanf("%s", str + 1);
  int n = strlen(str + 1);
  for (int i = 1; i <= n; i++) a[i] = str[i] - '0';
  for (int i = 1; i <= 9; i++) f[0][0][i][0] = 1;
  for (int i = 0; i <= n - 1; i++)
    for (int j = 0; j <= i; j++)
      for (int k = 1; k <= 9; k++)
        for (int l = 0; l <= 1; l++)
          if (f[i][j][k][l])
            for (int v = !l ? a[i + 1] : 9; v >= 0; v--)
              ad(f[i + 1][j + (v >= k)][k][l | (v < a[i + 1])], f[i][j][k][l]);
  long long p = 1, ans = 0;
  for (int i = 1; i <= n; i++) {
    long long sum = 0;
    for (int j = 1; j <= 9; j++) sum += f[n][i][j][0] + f[n][i][j][1];
    sum %= mod;
    ans = (ans + sum * p) % mod;
    p = (p * 10 + 1) % mod;
  }
  pr2(ans);
}
int main() {
  int T = 1;
  while (T--) solve();
  return 0;
}