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>
  constexpr int Block = 1e6;
  constexpr int Mod[] = {998244353, 1000000007};
  long long n, cnt;
  std::vector<int> val[7];
  int res, sum[7];
  void DFS1(int d, int v) {
    val[d].emplace_back((++cnt - v + Mod[0]) % Mod[0]);
    if (d == 6) return;
    for (int c = 0; c < 10; ++c) DFS1(d + 1, v * 10 + c);
  }
  void DFS2(int d, long long v) {
    if (v > n) return;
    if (d != 0) {
      if (v * Block > n / 10 && (v + 1) * Block - 1 <= n) {
        for (int i = 0, p = 1; i <= 6; ++i, p *= 10) {
          int coef = (cnt - v * p % Mod[0] + Mod[0]) % Mod[0];
          auto iter =
              std::lower_bound(val[i].begin(), val[i].end(), Mod[0] - coef);
          res = (res + 1LL * coef * val[i].size() + sum[i]) % Mod[1];
          res = (res - 1LL * Mod[0] * (val[i].end() - iter) % Mod[1] + Mod[1]) %
                Mod[1];
        }
        for (int i = 0; i <= 6; ++i) cnt += val[i].size();
        return;
      }
      res = (res + (++cnt - v % Mod[0] + Mod[0]) % Mod[0]) % Mod[1];
    }
    for (int c = !d; c < 10; ++c) DFS2(d + 1, v * 10 + c);
  }
  int main() {
    scanf("%lld", &n);
    DFS1(0, 0);
    for (int d = 0; d <= 6; ++d) {
      std::sort(val[d].begin(), val[d].end());
      for (int x : val[d]) sum[d] = (sum[d] + x) % Mod[1];
    }
    cnt = 0;
    DFS2(0, 0);
    printf("%d\n", res);
  }
  