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>
namespace FastIO {
bool IOerror = 0;
inline char nc() {
  static char buf[100000], *p1 = buf + 100000, *pend = buf + 100000;
  if (p1 == pend) {
    p1 = buf;
    pend = buf + fread(buf, 1, 100000, stdin);
    if (pend == p1) {
      IOerror = 1;
      return -1;
    }
  }
  return *p1++;
}
inline bool blank(char ch) {
  return ch == ' ' || ch == '\n' || ch == '\r' || ch == '\t';
}
template <class T>
inline bool read(T &x) {
  bool sign = 0;
  char ch = nc();
  x = 0;
  for (; blank(ch); ch = nc())
    ;
  if (IOerror) return false;
  if (ch == '-') sign = 1, ch = nc();
  for (; ch >= '0' && ch <= '9'; ch = nc()) x = x * 10 + ch - '0';
  if (sign) x = -x;
  return true;
}
template <class T, class... U>
bool read(T &h, U &...t) {
  return read(h) && read(t...);
}
};  // namespace FastIO
using namespace std;
using namespace FastIO;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const long long INF = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
long long n;
long long dp[710][710][10][2];
char s[710];
long long a[710];
signed main() {
  cin >> s + 1;
  n = strlen(s + 1);
  for (long long i = 1; i <= n; i++) a[i] = s[i] - '0';
  for (long long i = 0; i <= 9; i++) dp[0][0][i][1] = 1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j <= i; j++) {
      for (long long k = 1; k <= 9; k++) {
        for (long long lim = 0; lim <= 1; lim++) {
          long long up = lim ? a[i + 1] : 9;
          for (long long p = 0; p <= up; p++) {
            long long &x = dp[i + 1][j + (p >= k)][k][(lim && p == up)];
            x = (x + dp[i][j][k][lim]) % mod;
          }
        }
      }
    }
  }
  long long ans = 0;
  for (long long k = 1; k <= 9; k++) {
    long long sum = 1;
    for (long long j = 1; j <= n; j++) {
      ans = (ans + sum * (dp[n][j][k][0] + dp[n][j][k][1]) % mod) % mod;
      sum = (sum * 10 + 1) % mod;
    }
  }
  cout << ans << endl;
  return 0;
}