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;
  inline long long read() {
    char c = getchar();
    long long x = 0;
    bool f = 0;
    for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
    for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    if (f) x = -x;
    return x;
  }
  struct modint {
    long long x;
    modint(long long o = 0) { x = o; }
    modint &operator=(long long o) { return x = o, *this; }
    modint &operator+=(modint o) {
      return x = x + o.x >= 1000000007 ? x + o.x - 1000000007 : x + o.x, *this;
    }
    modint &operator-=(modint o) {
      return x = x - o.x < 0 ? x - o.x + 1000000007 : x - o.x, *this;
    }
    modint &operator*=(modint o) { return x = 1ll * x * o.x % 1000000007, *this; }
    modint &operator^=(long long b) {
      modint a = *this, c = 1;
      for (; b; b >>= 1, a *= a)
        if (b & 1) c *= a;
      return x = c.x, *this;
    }
    modint &operator/=(modint o) { return *this *= o ^= 1000000007 - 2; }
    modint &operator+=(long long o) {
      return x = x + o >= 1000000007 ? x + o - 1000000007 : x + o, *this;
    }
    modint &operator-=(long long o) {
      return x = x - o < 0 ? x - o + 1000000007 : x - o, *this;
    }
    modint &operator*=(long long o) {
      return x = 1ll * x * o % 1000000007, *this;
    }
    modint &operator/=(long long o) {
      return *this *= ((modint(o)) ^= 1000000007 - 2);
    }
    template <class I>
    friend modint operator+(modint a, I b) {
      return a += b;
    }
    template <class I>
    friend modint operator-(modint a, I b) {
      return a -= b;
    }
    template <class I>
    friend modint operator*(modint a, I b) {
      return a *= b;
    }
    template <class I>
    friend modint operator/(modint a, I b) {
      return a /= b;
    }
    friend modint operator^(modint a, long long b) { return a ^= b; }
    friend bool operator==(modint a, long long b) { return a.x == b; }
    friend bool operator!=(modint a, long long b) { return a.x != b; }
    bool operator!() { return !x; }
    modint operator-() { return x ? 1000000007 - x : 0; }
  };
  vector<modint> fac, ifac, iv;
  inline void initC(long long n) {
    if (iv.empty()) fac = ifac = iv = vector<modint>(2, 1);
    long long m = iv.size();
    ++n;
    if (m >= n) return;
    iv.resize(n), fac.resize(n), ifac.resize(n);
    for (register long long i = (m); i <= (n - 1); ++i) {
      iv[i] = iv[1000000007 % i] * (1000000007 - 1000000007 / i);
      fac[i] = fac[i - 1] * i, ifac[i] = ifac[i - 1] * iv[i];
    }
  }
  inline modint C(long long n, long long m) {
    if (m < 0 || n < m) return 0;
    return initC(n), fac[n] * ifac[m] * ifac[n - m];
  }
  inline modint sign(long long n) { return (n & 1) ? (1000000007 - 1) : (1); }
  inline long long sub(long long x, long long y) {
    return ((x - y) % 998244353 + 998244353) % 998244353;
  }
  long long n, rnk;
  vector<long long> v[13];
  modint sum[13];
  void pre(long long d, long long val) {
    ++rnk;
    v[d].push_back(sub(rnk, val));
    if (d == 6) return;
    for (register long long i = (0); i <= (9); ++i) pre(d + 1, val * 10 + i);
  }
  modint res = 0;
  modint calc(long long d, long long coef) {
    long long p =
        lower_bound(v[d].begin(), v[d].end(), 998244353 - coef) - v[d].begin();
    return (coef * v[d].size() % 1000000007 + sum[d] -
            1ll * 998244353 * (v[d].size() - p) % 1000000007);
  }
  void dfs(long long d, long long val) {
    if (val > n) return;
    if (d >= 1) {
      if (val * 10000000 > n && (val + 1) * 1000000 - 1 <= n) {
        long long pw10 = 1;
        for (register long long i = (0); i <= (6); ++i)
          res += calc(i, sub(rnk, val * pw10)), pw10 *= 10;
        for (register long long i = (0); i <= (6); ++i) rnk += v[i].size();
        return;
      }
      ++rnk;
      res += sub(rnk, val);
    }
    for (register long long i = ((!d)); i <= (9); ++i) dfs(d + 1, val * 10 + i);
  }
  signed main() {
    n = read();
    pre(0, 0);
    for (register long long d = (0); d <= (6); ++d) {
      sort(v[d].begin(), v[d].end());
      for (auto t : v[d]) sum[d] += t;
    }
    rnk = 0;
    dfs(0, 0);
    cout << res.x;
    return 0;
  }
  