#include <bits/stdc++.h>
using namespace std;
const int P = 998244353, mod = 1e9 + 7, B = 1e6;
int m(long long d) { return (d % P + P) % P; }
long long n, cnt, s[7];
int ans;
vector<int> p[7];
void dfs0(int d, long long v) {
  cnt++, p[d].push_back(m(cnt - v));
  if (d < 6)
    for (int i = 0; i <= 9; i++) dfs0(d + 1, v * 10 + i);
}
int calc(int id, int f) {
  int ps = lower_bound(p[id].begin(), p[id].end(), P - f) - p[id].begin();
  return ((long long)f * p[id].size() + s[id] -
          (long long)P * (p[id].size() - ps)) %
         mod;
}
void dfs1(int d, long long v) {
  if (v > n) return;
  if (d >= 1) {
    if (v * B > n / 10 && v * B + B - 1 <= n) {
      int t = 1;
      for (int i = 0; i <= 6; i++)
        (ans += calc(i, m(cnt - v * t))) %= mod, t *= 10;
      for (int i = 0; i <= 6; i++) cnt += p[i].size();
      return;
    }
    (ans += m(++cnt - v)) %= mod;
  }
  for (int i = !d; i <= 9; i++) dfs1(d + 1, v * 10 + i);
}
int main() {
  scanf("%lld", &n), dfs0(0, 0);
  for (int i = 0; i <= 6; i++) {
    sort(p[i].begin(), p[i].end());
    for (int j = 0; j <= p[i].size() - 1; j++) s[i] += p[i][j];
  }
  cnt = 0, dfs1(0, 0), printf("%d\n", ans);
  return 0;
}

### Prompt

Your task is to create a CPP solution to the following problem:

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;
const int P = 998244353, mod = 1e9 + 7, B = 1e6;
int m(long long d) { return (d % P + P) % P; }
long long n, cnt, s[7];
int ans;
vector<int> p[7];
void dfs0(int d, long long v) {
  cnt++, p[d].push_back(m(cnt - v));
  if (d < 6)
    for (int i = 0; i <= 9; i++) dfs0(d + 1, v * 10 + i);
}
int calc(int id, int f) {
  int ps = lower_bound(p[id].begin(), p[id].end(), P - f) - p[id].begin();
  return ((long long)f * p[id].size() + s[id] -
          (long long)P * (p[id].size() - ps)) %
         mod;
}
void dfs1(int d, long long v) {
  if (v > n) return;
  if (d >= 1) {
    if (v * B > n / 10 && v * B + B - 1 <= n) {
      int t = 1;
      for (int i = 0; i <= 6; i++)
        (ans += calc(i, m(cnt - v * t))) %= mod, t *= 10;
      for (int i = 0; i <= 6; i++) cnt += p[i].size();
      return;
    }
    (ans += m(++cnt - v)) %= mod;
  }
  for (int i = !d; i <= 9; i++) dfs1(d + 1, v * 10 + i);
}
int main() {
  scanf("%lld", &n), dfs0(0, 0);
  for (int i = 0; i <= 6; i++) {
    sort(p[i].begin(), p[i].end());
    for (int j = 0; j <= p[i].size() - 1; j++) s[i] += p[i][j];
  }
  cnt = 0, dfs1(0, 0), printf("%d\n", ans);
  return 0;
}

```