#include <bits/stdc++.h>
using namespace std;
long long a, b, mod1 = 998244353, mod2 = 1000000007, rk, an, sum[101];
vector<long long> qu[101];
void dfs1(long long qq, long long ww) {
  if (qq > 6) return;
  rk++;
  qu[qq].push_back(((rk - ww) % mod1 + mod1) % mod1);
  sum[qq] += ((rk - ww) % mod1 + mod1) % mod1;
  for (int i = 0; i <= 9; i++) {
    dfs1(qq + 1, ww * 10 + i);
  }
}
long long work(long long qq, long long ww) {
  long long tt =
      lower_bound(qu[qq].begin(), qu[qq].end(), mod1 - ww) - qu[qq].begin();
  return (ww * qu[qq].size() % mod2 + sum[qq] -
          mod1 * (qu[qq].size() - tt) % mod2) %
         mod2;
}
void dfs2(long long qq, long long ww) {
  if (ww > a) return;
  if (qq >= 1) {
    if (ww * 10000000 > a && ww * 1000000 + 999999 <= a) {
      long long tt = 1;
      for (int i = 0; i <= 6; i++) {
        an += work(i, ((rk - ww * tt) % mod1 + mod1) % mod1);
        an %= mod2;
        tt *= 10;
      }
      for (int i = 0; i <= 6; i++) {
        rk += qu[i].size();
      }
      return;
    }
    rk++;
    an += ((rk - ww) % mod1 + mod1) % mod1;
    an %= mod2;
  }
  if (qq == 0)
    for (int i = 1; i <= 9; i++) dfs2(qq + 1, ww * 10 + i);
  else
    for (int i = 0; i <= 9; i++) dfs2(qq + 1, ww * 10 + i);
}
int main() {
  scanf("%lld", &a);
  dfs1(0, 0);
  rk = 0;
  for (int i = 0; i <= 6; i++) sort(qu[i].begin(), qu[i].end());
  dfs2(0, 0);
  an %= mod2;
  an += mod2;
  an %= mod2;
  printf("%lld", an);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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>
using namespace std;
long long a, b, mod1 = 998244353, mod2 = 1000000007, rk, an, sum[101];
vector<long long> qu[101];
void dfs1(long long qq, long long ww) {
  if (qq > 6) return;
  rk++;
  qu[qq].push_back(((rk - ww) % mod1 + mod1) % mod1);
  sum[qq] += ((rk - ww) % mod1 + mod1) % mod1;
  for (int i = 0; i <= 9; i++) {
    dfs1(qq + 1, ww * 10 + i);
  }
}
long long work(long long qq, long long ww) {
  long long tt =
      lower_bound(qu[qq].begin(), qu[qq].end(), mod1 - ww) - qu[qq].begin();
  return (ww * qu[qq].size() % mod2 + sum[qq] -
          mod1 * (qu[qq].size() - tt) % mod2) %
         mod2;
}
void dfs2(long long qq, long long ww) {
  if (ww > a) return;
  if (qq >= 1) {
    if (ww * 10000000 > a && ww * 1000000 + 999999 <= a) {
      long long tt = 1;
      for (int i = 0; i <= 6; i++) {
        an += work(i, ((rk - ww * tt) % mod1 + mod1) % mod1);
        an %= mod2;
        tt *= 10;
      }
      for (int i = 0; i <= 6; i++) {
        rk += qu[i].size();
      }
      return;
    }
    rk++;
    an += ((rk - ww) % mod1 + mod1) % mod1;
    an %= mod2;
  }
  if (qq == 0)
    for (int i = 1; i <= 9; i++) dfs2(qq + 1, ww * 10 + i);
  else
    for (int i = 0; i <= 9; i++) dfs2(qq + 1, ww * 10 + i);
}
int main() {
  scanf("%lld", &a);
  dfs1(0, 0);
  rk = 0;
  for (int i = 0; i <= 6; i++) sort(qu[i].begin(), qu[i].end());
  dfs2(0, 0);
  an %= mod2;
  an += mod2;
  an %= mod2;
  printf("%lld", an);
  return 0;
}

```