#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007, MOD = 998244353, maxk = 7, base = 1000000;
long long n, ord;
int ans;
int sum[maxk];
vector<int> v[maxk];
void dfs1(int x, int len) {
  ord++, v[len].push_back(((ord - x) % MOD + MOD) % MOD),
      sum[len] = (sum[len] + v[len].back()) % mod;
  if (len == 6) return;
  for (int i = 0; i <= 9; i++) dfs1(x * 10 + i, len + 1);
}
void dfs2(long long x) {
  if (x > n) return;
  if (x * base + base - 1 <= n && x * base * 10 > n) {
    long long now = ord;
    for (int i = 0, mul = 1; i <= 6; i++, mul *= 10) {
      int sz = v[i].size(), dif = ((now - x * mul) % MOD + MOD) % MOD,
          pos = lower_bound(v[i].begin(), v[i].end(), MOD - dif) - v[i].begin();
      ans = (ans + 1ll * dif * sz + sum[i]) % mod,
      ans = (ans - 1ll * MOD * (sz - pos) % mod + mod) % mod;
      ord += sz;
    }
    return;
  }
  ord++, ans = (ans + ((ord - x) % MOD + MOD) % MOD) % mod;
  for (int i = 0; i <= 9; i++) dfs2(x * 10 + i);
}
int main() {
  scanf("%lld", &n);
  ord = 0, dfs1(0, 0);
  for (int i = 0; i <= 6; i++) sort(v[i].begin(), v[i].end());
  ord = 0;
  for (int i = 1; i <= 9; i++) dfs2(i);
  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>
using namespace std;
const int mod = 1000000007, MOD = 998244353, maxk = 7, base = 1000000;
long long n, ord;
int ans;
int sum[maxk];
vector<int> v[maxk];
void dfs1(int x, int len) {
  ord++, v[len].push_back(((ord - x) % MOD + MOD) % MOD),
      sum[len] = (sum[len] + v[len].back()) % mod;
  if (len == 6) return;
  for (int i = 0; i <= 9; i++) dfs1(x * 10 + i, len + 1);
}
void dfs2(long long x) {
  if (x > n) return;
  if (x * base + base - 1 <= n && x * base * 10 > n) {
    long long now = ord;
    for (int i = 0, mul = 1; i <= 6; i++, mul *= 10) {
      int sz = v[i].size(), dif = ((now - x * mul) % MOD + MOD) % MOD,
          pos = lower_bound(v[i].begin(), v[i].end(), MOD - dif) - v[i].begin();
      ans = (ans + 1ll * dif * sz + sum[i]) % mod,
      ans = (ans - 1ll * MOD * (sz - pos) % mod + mod) % mod;
      ord += sz;
    }
    return;
  }
  ord++, ans = (ans + ((ord - x) % MOD + MOD) % MOD) % mod;
  for (int i = 0; i <= 9; i++) dfs2(x * 10 + i);
}
int main() {
  scanf("%lld", &n);
  ord = 0, dfs1(0, 0);
  for (int i = 0; i <= 6; i++) sort(v[i].begin(), v[i].end());
  ord = 0;
  for (int i = 1; i <= 9; i++) dfs2(i);
  printf("%d\n", ans);
  return 0;
}

```