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 = 700 + 5;
template <typename T>
bool cmax(T &a, T b) {
  return (a < b) ? a = b, 1 : 0;
}
template <typename T>
bool cmin(T &a, T b) {
  return (a > b) ? a = b, 1 : 0;
}
template <typename T>
T read() {
  T ans = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch) && ch != '-') ch = getchar();
  if (ch == '-') f = -1, ch = getchar();
  while (isdigit(ch))
    ans = (ans << 3) + (ans << 1) + (ch - '0'), ch = getchar();
  return ans * f;
}
template <typename T>
void write(T x, char y) {
  if (x == 0) {
    putchar('0'), putchar(y);
    return;
  }
  if (x < 0) {
    putchar('-');
    x = -x;
  }
  static char wr[20];
  int top = 0;
  for (; x; x /= 10) wr[++top] = x % 10 + '0';
  while (top) putchar(wr[top--]);
  putchar(y);
}
void file() {}
int dp[N][N][10][2], n;
char s[N];
void input() {
  scanf("%s", s + 1);
  n = strlen(s + 1);
}
const int mo = 1e9 + 7;
int mc[N];
void add(int &x, int y) {
  x += y;
  x -= (x >= mo ? mo : 0);
}
void work() {
  int limit;
  mc[1] = 1;
  for (register int i = (int)(2); i <= (int)(n); ++i)
    mc[i] = 1ll * ((long long)mc[i - 1] * 10 + 1) % mo;
  for (register int i = (int)(0); i <= (int)(9); ++i) dp[0][0][i][1] = 1;
  for (register int i = (int)(0); i <= (int)(n - 1); ++i)
    for (register int j = (int)(0); j <= (int)(i); ++j)
      for (register int k = (int)(0); k <= (int)(9); ++k)
        for (register int o = (int)(0); o <= (int)(1); ++o)
          if (dp[i][j][k][o]) {
            limit = o ? s[i + 1] - '0' : 9;
            for (register int now = (int)(0); now <= (int)(limit); ++now) {
              add(dp[i + 1][j + (now >= k)][k][o & (now == limit)],
                  dp[i][j][k][o]);
            }
          }
  int ans = 0;
  for (register int i = (int)(1); i <= (int)(n); ++i)
    for (register int k = (int)(1); k <= (int)(9); ++k)
      add(ans, 1ll * mc[i] * (dp[n][i][k][0] + dp[n][i][k][1]) % mo);
  write(ans, '\n');
}
int main() {
  file();
  input();
  work();
  reverse(s + 1, s + n + 1);
  return 0;
}