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;
  const int MOD = 1e9 + 7;
  const int B = 1000000;
  int xmod(long long d) {
    int ret = d % P;
    return ret < 0 ? ret + P : ret;
  }
  long long n, cnt;
  int ans;
  vector<int> hv[7];
  long long sum[7];
  void dfs0(int dep, long long val) {
    cnt++;
    hv[dep].push_back(xmod(cnt - val));
    if (dep == 6) return;
    for (int i = 0; i < (int)(10); i++) dfs0(dep + 1, val * 10 + i);
  }
  int calc(int id, int coef) {
    int pos =
        lower_bound(hv[id].begin(), hv[id].end(), P - coef) - hv[id].begin();
    return (1LL * coef * (int)hv[id].size() + sum[id] -
            1LL * P * ((int)hv[id].size() - pos)) %
           MOD;
  }
  void dfs1(int dep, long long val) {
    if (val > n) return;
    if (dep >= 1) {
      if (val * B > n / 10 && val * B + B - 1 <= n) {
        int tmp = 1;
        for (int i = 0; i < (int)(7); i++) {
          ans = (ans + calc(i, xmod(cnt - val * tmp))) % MOD;
          tmp *= 10;
        }
        for (int i = 0; i < (int)(7); i++) cnt += hv[i].size();
        return;
      }
      cnt++;
      ans = (ans + xmod(cnt - val)) % MOD;
    }
    for (int i = !dep; i < 10; i++) dfs1(dep + 1, val * 10 + i);
  }
  int main() {
    scanf("%lld", &n);
    dfs0(0, 0);
    for (int i = 0; i < (int)(7); i++) {
      sort(hv[i].begin(), hv[i].end());
      for (int j = 0; j < (int)(hv[i].size()); j++) sum[i] += hv[i][j];
    }
    cnt = 0;
    dfs1(0, 0);
    printf("%d\n", ans);
    return 0;
  }
  