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;
namespace IO {
const int sz = 1 << 15;
char inbuf[sz], outbuf[sz];
char *pinbuf = inbuf + sz;
char *poutbuf = outbuf;
inline char _getchar() {
  if (pinbuf == inbuf + sz) fread(inbuf, 1, sz, stdin), pinbuf = inbuf;
  return *(pinbuf++);
}
inline void _putchar(char x) {
  if (poutbuf == outbuf + sz) fwrite(outbuf, 1, sz, stdout), poutbuf = outbuf;
  *(poutbuf++) = x;
}
inline void flush() {
  if (poutbuf != outbuf)
    fwrite(outbuf, 1, poutbuf - outbuf, stdout), poutbuf = outbuf;
}
}  // namespace IO
inline int read() {
  int x = 0, p = 1;
  char c = IO::_getchar();
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = IO::_getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = IO::_getchar();
  return x * p;
}
namespace Mymath {
long long qp(long long x, long long p, long long mod) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = ans * x % mod;
    x = x * x % mod;
    p >>= 1;
  }
  return ans;
}
long long inv(long long x, long long mod) { return qp(x, mod - 2, mod); }
long long C(long long N, long long K, long long fact[], long long mod) {
  return fact[N] * inv(fact[K], mod) % mod * inv(fact[N - K], mod) % mod;
}
template <typename Tp>
Tp gcd(Tp A, Tp B) {
  if (B == 0) return A;
  return gcd(B, A % B);
}
template <typename Tp>
Tp lcm(Tp A, Tp B) {
  return A * B / gcd(A, B);
}
};  // namespace Mymath
const int Maxn = 705;
const long long mod = 1e9 + 7;
int dp[Maxn][3][Maxn][11];
int tot[11][Maxn];
char c[Maxn];
int n;
void Add(int &x, int y) {
  x += y;
  if (x >= mod) x -= mod;
}
int main() {
  scanf("%s", c);
  n = strlen(c);
  for (int i = 0; i <= 9; i++) dp[0][1][0][i] = 1;
  for (int i = 0; i < n; i++) {
    int curv = c[i] - '0';
    for (int j = 0; j <= i; j++) {
      for (int k = 0; k <= 9; k++) {
        if (i)
          for (int l = 0; l < 3; l++) Add(tot[k][j], dp[i][l][j][k]);
        Add(dp[i + 1][1][j + (curv >= k)][k], dp[i][1][j][k]);
        for (int l = 0; l <= 9; l++) {
          if (!i && !l) continue;
          if (l == curv) continue;
          Add(dp[i + 1][(l < curv) ? 0 : 2][j + (l >= k)][k], dp[i][1][j][k]);
        }
        for (int l = 0; l <= 9; l++) {
          if (!i && !l) continue;
          Add(dp[i + 1][0][j + (l >= k)][k], dp[i][0][j][k]);
          Add(dp[i + 1][2][j + (l >= k)][k], dp[i][2][j][k]);
        }
      }
    }
  }
  for (int j = 0; j <= n; j++) {
    for (int k = 0; k <= 9; k++) {
      for (int l = 0; l < 2; l++) Add(tot[k][j], dp[n][l][j][k]);
    }
  }
  for (int i = 0; i <= 9; i++) {
    for (int j = n; j >= 0; j--) {
      Add(tot[i][j], tot[i][j + 1]);
    }
  }
  long long ans = 0;
  int w = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 1; j <= 9; j++) {
      int cnt = tot[j][i + 1] - tot[j + 1][i + 1];
      if (cnt < 0) cnt += mod;
      ans += 1ll * cnt * w % mod * j;
      ans %= mod;
    }
    w = 1ll * w * 10 % mod;
  }
  printf("%I64d\n", ans);
  return 0;
}