#include <bits/stdc++.h>
template <typename _Tp>
void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
const int B = 1e6, mod1 = 998244353, mod2 = 1000000007;
std::vector<int> v[7];
long long sum[7], cnt, n;
int ans;
inline int norm(long long x) {
  x %= mod1;
  return x < 0 ? x + mod1 : x;
}
void dfs1(int dep, int val) {
  ++cnt, v[dep].push_back(norm(cnt - val));
  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 (val * B > n / 10 && (val + 1) * B - 1 <= n) {
      long long tmp = val;
      for (int i = 0; i <= 6; ++i) {
        int cur = norm(cnt - tmp);
        ans = (ans + 1LL * cur * ((int)v[i].size()) + sum[i] -
               1LL * mod1 *
                   (v[i].end() -
                    std::lower_bound(v[i].begin(), v[i].end(), mod1 - cur))) %
              mod2;
        tmp *= 10;
      }
      for (int i = 0; i <= 6; ++i) cnt += ((int)v[i].size());
      return;
    }
    ++cnt, ans = (ans + norm(cnt - val)) % mod2;
  }
  for (int i = !dep; i < 10; ++i) dfs2(dep + 1, 10LL * val + i);
}
int main() {
  read(n), dfs1(0, 0);
  for (int i = 0; i <= 6; ++i) {
    std::sort(v[i].begin(), v[i].end());
    for (auto it : v[i]) sum[i] += it;
  }
  cnt = 0, dfs2(0, 0);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
template <typename _Tp>
void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
const int B = 1e6, mod1 = 998244353, mod2 = 1000000007;
std::vector<int> v[7];
long long sum[7], cnt, n;
int ans;
inline int norm(long long x) {
  x %= mod1;
  return x < 0 ? x + mod1 : x;
}
void dfs1(int dep, int val) {
  ++cnt, v[dep].push_back(norm(cnt - val));
  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 (val * B > n / 10 && (val + 1) * B - 1 <= n) {
      long long tmp = val;
      for (int i = 0; i <= 6; ++i) {
        int cur = norm(cnt - tmp);
        ans = (ans + 1LL * cur * ((int)v[i].size()) + sum[i] -
               1LL * mod1 *
                   (v[i].end() -
                    std::lower_bound(v[i].begin(), v[i].end(), mod1 - cur))) %
              mod2;
        tmp *= 10;
      }
      for (int i = 0; i <= 6; ++i) cnt += ((int)v[i].size());
      return;
    }
    ++cnt, ans = (ans + norm(cnt - val)) % mod2;
  }
  for (int i = !dep; i < 10; ++i) dfs2(dep + 1, 10LL * val + i);
}
int main() {
  read(n), dfs1(0, 0);
  for (int i = 0; i <= 6; ++i) {
    std::sort(v[i].begin(), v[i].end());
    for (auto it : v[i]) sum[i] += it;
  }
  cnt = 0, dfs2(0, 0);
  printf("%d\n", ans);
  return 0;
}

```