description: Integers from 1 to n (inclusive) were sorted lexicographically (considering integers as strings). As a result, array a_1, a_2, ..., a_n was obtained.
  
  Calculate value of (∑_{i = 1}^n ((i - a_i) mod 998244353)) mod 10^9 + 7.
  
  x mod y here means the remainder after division x by y. This remainder is always non-negative and doesn't exceed y - 1. For example, 5 mod 3 = 2, (-1) mod 6 = 5. 
  
  Input
  
  The first line contains the single integer n (1 ≤ n ≤ 10^{12}).
  
  Output
  
  Print one integer — the required sum.
  
  Examples
  
  Input
  
  
  3
  
  
  Output
  
  
  0
  
  
  Input
  
  
  12
  
  
  Output
  
  
  994733045
  
  
  Input
  
  
  21
  
  
  Output
  
  
  978932159
  
  
  Input
  
  
  1000000000000
  
  
  Output
  
  
  289817887
  
  Note
  
  A string a is lexicographically smaller than a string b if and only if one of the following holds:
  
    * a is a prefix of b, but a ≠ b; 
    * in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b. 
  
  
  
  For example, 42 is lexicographically smaller than 6, because they differ in the first digit, and 4 < 6; 42 < 420, because 42 is a prefix of 420.
  
  Let's denote 998244353 as M.
  
  In the first example, array a is equal to [1, 2, 3]. 
  
    * (1 - 1) mod M = 0 mod M = 0 
    * (2 - 2) mod M = 0 mod M = 0 
    * (3 - 3) mod M = 0 mod M = 0 
  
  
  
  As a result, (0 + 0 + 0) mod 10^9 + 7 = 0
  
  In the second example, array a is equal to [1, 10, 11, 12, 2, 3, 4, 5, 6, 7, 8, 9]. 
  
    * (1 - 1) mod M = 0 mod M = 0 
    * (2 - 10) mod M = (-8) mod M = 998244345 
    * (3 - 11) mod M = (-8) mod M = 998244345 
    * (4 - 12) mod M = (-8) mod M = 998244345 
    * (5 - 2) mod M = 3 mod M = 3 
    * (6 - 3) mod M = 3 mod M = 3 
    * (7 - 4) mod M = 3 mod M = 3 
    * (8 - 5) mod M = 3 mod M = 3 
    * (9 - 6) mod M = 3 mod M = 3 
    * (10 - 7) mod M = 3 mod M = 3 
    * (11 - 8) mod M = 3 mod M = 3 
    * (12 - 9) mod M = 3 mod M = 3 
  
  
  
  As a result, (0 + 998244345 + 998244345 + 998244345 + 3 + 3 + 3 + 3 + 3 + 3 + 3 + 3) mod 10^9 + 7 = 2994733059 mod 10^9 + 7 = 994733045
solution: #include <bits/stdc++.h>
  using namespace std;
  long long a, b, mod1 = 998244353, mod2 = 1000000007, rk, an, sum[101];
  vector<long long> qu[101];
  void dfs1(long long qq, long long ww) {
    if (qq > 6) return;
    rk++;
    qu[qq].push_back(((rk - ww) % mod1 + mod1) % mod1);
    sum[qq] += ((rk - ww) % mod1 + mod1) % mod1;
    for (int i = 0; i <= 9; i++) {
      dfs1(qq + 1, ww * 10 + i);
    }
  }
  long long work(long long qq, long long ww) {
    long long tt =
        lower_bound(qu[qq].begin(), qu[qq].end(), mod1 - ww) - qu[qq].begin();
    return (ww * qu[qq].size() % mod2 + sum[qq] -
            mod1 * (qu[qq].size() - tt) % mod2) %
           mod2;
  }
  void dfs2(long long qq, long long ww) {
    if (ww > a) return;
    if (qq >= 1) {
      if (ww * 10000000 > a && ww * 1000000 + 999999 <= a) {
        long long tt = 1;
        for (int i = 0; i <= 6; i++) {
          an += work(i, ((rk - ww * tt) % mod1 + mod1) % mod1);
          an %= mod2;
          tt *= 10;
        }
        for (int i = 0; i <= 6; i++) {
          rk += qu[i].size();
        }
        return;
      }
      rk++;
      an += ((rk - ww) % mod1 + mod1) % mod1;
      an %= mod2;
    }
    if (qq == 0)
      for (int i = 1; i <= 9; i++) dfs2(qq + 1, ww * 10 + i);
    else
      for (int i = 0; i <= 9; i++) dfs2(qq + 1, ww * 10 + i);
  }
  int main() {
    scanf("%lld", &a);
    dfs1(0, 0);
    rk = 0;
    for (int i = 0; i <= 6; i++) sort(qu[i].begin(), qu[i].end());
    dfs2(0, 0);
    an %= mod2;
    an += mod2;
    an %= mod2;
    printf("%lld", an);
    return 0;
  }
  