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 N = 1e6, M = N * 2 + 10, K = 6, P1 = 998244353, P2 = 1e9 + 7;
  long long n;
  int ans = 0;
  int del[K + 1][M], sum[K + 1][M], tot[K + 1];
  void Add(int &first, int second) {
    if ((first += second) >= P2) first -= P2;
  }
  void Sub(int &first, int second) {
    if ((first -= second) < 0) first += P2;
  }
  string To_string(int first, int second) {
    string num;
    for (int i = 0; i < second; i++, first /= 10) {
      num.push_back(first % 10);
    }
    reverse(num.begin(), num.end());
    return num;
  }
  void dfs1(int now, int len, int &rank) {
    if (len > K) return;
    rank++;
    del[len][++tot[len]] = (rank - now + P1) % P1;
    for (int i = 0; i < 10; i++) dfs1(now * 10 + i, len + 1, rank);
    return;
  }
  void dfs2(long long now, int &rank) {
    if (now > n) return;
    if (now * N * 10 > n && now * N + N - 1 <= n) {
      int p10 = 1, res = rank;
      for (int i = 0; i <= K; i++) {
        int d = (rank - 1ll * now * p10 % P1 + P1) % P1;
        int p = lower_bound(del[i] + 1, del[i] + tot[i] + 1, P1 - d) - del[i];
        Add(ans, 1ll * tot[i] * d % P2);
        Add(ans, sum[i][tot[i]]);
        Sub(ans, 1ll * (tot[i] - p + 1) * P1 % P2);
        (res += p10) %= P1;
        p10 *= 10;
      }
      rank = res;
    } else {
      rank = (rank + 1) % P1;
      Add(ans, (rank - now % P1 + P1) % P1);
      for (int i = 0; i < 10; i++) dfs2(now * 10 + i, rank);
    }
    return;
  }
  int main() {
    scanf("%lld", &n);
    int cnt = 0;
    dfs1(0, 0, cnt);
    for (int i = 0; i <= K; i++) {
      sort(del[i] + 1, del[i] + tot[i] + 1);
      for (int j = 1; j <= tot[i]; j++)
        sum[i][j] = (sum[i][j - 1] + del[i][j]) % P2;
    }
    cnt = 0;
    for (int i = 1; i < 10; i++) dfs2(i, cnt);
    printf("%d\n", ans);
    return 0;
  }
  