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;
  long long n;
  int ans = 0;
  int digcnt;
  long long pref[20];
  long long pwr10[20];
  long long count_pref(long long x, int digits) {
    if (digits == 0) return n;
    long long Ans = 0;
    for (int i = digits; i < digcnt; i++) {
      Ans += pwr10[i - digits];
    }
    if (x < pref[digits]) {
      Ans += pwr10[digcnt - digits];
    } else if (x == pref[digits]) {
      Ans += n - x * pwr10[digcnt - digits] + 1;
    }
    return Ans;
  }
  void dfs(long long x, int digits, int upto, long long presum) {
    if (upto == digcnt) {
      if (x > pref[digits]) return;
      if (x == pref[digits] && digits < upto) {
        long long cursum = presum;
        for (long long nxt = pref[digits] * 10LL; nxt <= pref[digits + 1];
             nxt++) {
          if (digits || nxt) {
            dfs(nxt, digits + 1, upto, cursum + 1);
            cursum += count_pref(nxt, digits + 1);
          }
        }
        return;
      }
    }
    if (digits == upto) {
      ans -= (((presum + 1LL - x) + 1000000LL * 998244353LL) / 998244353LL -
              1000000LL);
      if (ans < 0) ans += 1000000007;
      if (ans >= 1000000007) ans -= 1000000007;
      return;
    }
    if (digits > 0) {
      long long mina = x * pwr10[upto - digits];
      long long maxa = mina + pwr10[upto - digits] - 1;
      long long mini = presum;
      long long maxi = presum + count_pref(x, digits);
      if ((((maxi - mina) + 1000000LL * 998244353LL) / 998244353LL - 1000000LL) ==
          (((mini - maxa) + 1000000LL * 998244353LL) / 998244353LL - 1000000LL)) {
        ans -= (((long long)((((maxi - mina) + 1000000LL * 998244353LL) /
                                  998244353LL -
                              1000000LL) +
                             1000000007) *
                 (long long)(pwr10[upto - digits] % 1000000007)) %
                1000000007);
        if (ans < 0) ans += 1000000007;
        if (ans >= 1000000007) ans -= 1000000007;
        return;
      }
    }
    long long cursum = presum;
    for (long long nxt = x * 10LL; nxt < (x + 1LL) * 10LL; nxt++) {
      if (digits || nxt) {
        dfs(nxt, digits + 1, upto, cursum + 1);
        cursum += count_pref(nxt, digits + 1);
      }
    }
  }
  int main() {
    pwr10[0] = 1LL;
    for (int i = 1; i <= 15; i++) {
      pwr10[i] = 10LL * pwr10[i - 1];
    }
    scanf("%lld", &n);
    char tmp[20];
    sprintf(tmp, "%lld", n);
    digcnt = strlen(tmp);
    long long tmp2 = n;
    for (int i = 0; i < digcnt; i++) {
      pref[digcnt - i] = tmp2;
      tmp2 /= 10LL;
    }
    for (int i = 1; i <= digcnt; i++) {
      dfs(0, 0, i, -1);
    }
    ans = (((long long)(ans) * (long long)(998244353)) % 1000000007);
    printf("%d", ans);
    return 0;
  }
  