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;
  const int P = 998244353, mod = 1e9 + 7, B = 1e6;
  int m(long long d) { return (d % P + P) % P; }
  long long n, cnt, s[7];
  int ans;
  vector<int> p[7];
  void dfs0(int d, long long v) {
    cnt++, p[d].push_back(m(cnt - v));
    if (d < 6)
      for (int i = 0; i <= 9; i++) dfs0(d + 1, v * 10 + i);
  }
  int calc(int id, int f) {
    int ps = lower_bound(p[id].begin(), p[id].end(), P - f) - p[id].begin();
    return ((long long)f * p[id].size() + s[id] -
            (long long)P * (p[id].size() - ps)) %
           mod;
  }
  void dfs1(int d, long long v) {
    if (v > n) return;
    if (d >= 1) {
      if (v * B > n / 10 && v * B + B - 1 <= n) {
        int t = 1;
        for (int i = 0; i <= 6; i++)
          (ans += calc(i, m(cnt - v * t))) %= mod, t *= 10;
        for (int i = 0; i <= 6; i++) cnt += p[i].size();
        return;
      }
      (ans += m(++cnt - v)) %= mod;
    }
    for (int i = !d; i <= 9; i++) dfs1(d + 1, v * 10 + i);
  }
  int main() {
    scanf("%lld", &n), dfs0(0, 0);
    for (int i = 0; i <= 6; i++) {
      sort(p[i].begin(), p[i].end());
      for (int j = 0; j <= p[i].size() - 1; j++) s[i] += p[i][j];
    }
    cnt = 0, dfs1(0, 0), printf("%d\n", ans);
    return 0;
  }
  