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;
  template <typename T>
  void read(T &t) {
    t = 0;
    char ch = getchar();
    int f = 1;
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    do {
      (t *= 10) += ch - '0';
      ch = getchar();
    } while ('0' <= ch && ch <= '9');
    t *= f;
  }
  template <typename T>
  void write(T t) {
    if (t < 0) {
      putchar('-');
      write(-t);
      return;
    }
    if (t > 9) write(t / 10);
    putchar('0' + t % 10);
  }
  template <typename T>
  void writeln(T t) {
    write(t);
    puts("");
  }
  const int mod = 998244353;
  const int mod2 = (1e9) + 7;
  long long n, cnt;
  int ans, sum[10];
  vector<int> vec[10];
  void update(int &x, int y) {
    x += y;
    if (x >= mod2) x -= mod2;
  }
  void dfs1(int dep, int val) {
    if (val > n) return;
    cnt++;
    vec[dep].push_back((cnt - val + mod) % mod);
    if (dep == 6) return;
    for (int i = 0; i < 10; i++) dfs1(dep + 1, val * 10 + i);
  }
  void dfs2(int dep, long long val) {
    if (val > n) return;
    if (dep >= 1) {
      if ((long long)val * 1000000 > n / 10 &&
          (long long)val * 1000000 + 999999 <= n) {
        long long tmp = val;
        for (int i = 0; i <= 6; i++) {
          int cur = (cnt % mod - tmp % mod + mod) % mod;
          update(ans, (long long)cur * (int)vec[i].size() % mod2);
          int pos =
              vec[i].end() - lower_bound(vec[i].begin(), vec[i].end(), mod - cur);
          update(ans, (sum[i] - (long long)pos * mod % mod2 + mod2) % mod2);
          tmp *= 10;
        }
        for (int i = 0; i <= 6; i++) cnt += (int)vec[i].size();
        return;
      }
      cnt++;
      update(ans, (cnt % mod - val % mod + mod) % mod);
    }
    for (int i = (dep == 0 ? 1 : 0); i < 10; i++)
      dfs2(dep + 1, (long long)val * 10 + i);
  }
  int main() {
    read(n);
    dfs1(0, 0);
    for (int i = 0; i <= 6; i++) {
      sort(vec[i].begin(), vec[i].end());
      for (int &x : vec[i]) update(sum[i], x);
    }
    cnt = 0;
    dfs2(0, 0);
    printf("%d\n", ans);
    return 0;
  }
  