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 = 1e5 + 10;
const long long mod = 1e9 + 7;
int n;
int a[N];
long long pow_m[N], c[N], v[N], fac[N], ifac[N];
long long f[12][N], g[N], h[12], n1[12], n2[12];
void Add(long long &x, long long y) { x = (x + y) % mod; }
long long qpow(long long x, long long y = mod - 2) {
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % mod;
    x = x * x % mod, y >>= 1;
  }
  return res;
}
long long C(long long x, long long y) {
  return fac[x] * ifac[y] % mod * ifac[x - y] % mod;
}
long long ans = 0;
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  pow_m[0] = 1;
  for (int i = 1; i < N; i++) pow_m[i] = pow_m[i - 1] * 10 % mod;
  fac[0] = 1;
  for (int i = 1; i < N; i++) fac[i] = fac[i - 1] * i % mod;
  ifac[N - 1] = qpow(fac[N - 1]);
  for (int i = N - 2; i >= 0; i--) ifac[i] = ifac[i + 1] * (i + 1) % mod;
  v[0] = 0;
  for (int i = 1; i < N; i++) v[i] = (v[i - 1] * 10 + 1) % mod;
  long long inv_9 = qpow(9);
  for (int i = 1; i <= 9; i++) {
    long long mof = 100 - 9 * i, mog = 10, nwf = 1, nwg = 1;
    f[i][0] = nwf, g[0] = nwg;
    for (int j = 1; j < N; j++) {
      nwf = (nwf * mof) % mod;
      nwg = (nwg * mog) % mod;
      f[i][j] = nwf * ifac[j] % mod;
      g[j] = nwg * ifac[j] % mod;
    }
  }
  string s;
  cin >> s;
  n = s.length();
  for (int i = 0; i < n; i++) a[n - i - 1] = s[i] - '0';
  for (int i = n - 1; i >= 0; i--) {
    h[9] = c[9];
    for (int j = 8; j >= 0; j--) {
      h[j] = h[j + 1] + c[j];
    }
    for (int j = 9; j >= 1; j--) {
      if (i == 0)
        n1[j] = min(a[i] + 1, j);
      else
        n1[j] = min(a[i], j);
      if (i == 0)
        n2[j] = max(0, a[i] - j + 1);
      else
        n2[j] = max(0, a[i] - j);
    }
    long long res = 0;
    for (int j = 1; j <= 9; j++) {
      Add(res, f[j][i] *
                   (pow_m[h[j]] * n1[j] % mod + pow_m[h[j] + 1] * n2[j] % mod) %
                   mod);
      Add(res, (n1[j] + n2[j]) * (mod - g[i]) % mod);
    }
    Add(ans, inv_9 * res % mod * fac[i] % mod);
    c[a[i]]++;
  }
  cout << ans << '\n';
  return 0;
}