#include <bits/stdc++.h>
using namespace std;
const int mod1 = 998244353, mod2 = 1e9 + 7;
const int N = 2e6, MX = 1e6;
long long n, c;
int tot, ans, lim, pw[N], sum[N];
vector<int> ve[N];
inline int dec(int x, int y, int mod) {
  return (x - y < 0) ? x - y + mod : x - y;
}
inline int add(int x, int y, int mod) {
  return (x + y >= mod) ? x + y - mod : x + y;
}
inline void dfs(int pos, int now) {
  if (now <= n) {
    ++tot;
    int x = dec(tot, now, mod1);
    ve[pos - 1].push_back(x);
    sum[pos - 1] = add(sum[pos - 1], x, mod2);
  }
  if (pos > 6) return;
  for (int i = 0; i <= 9; ++i) dfs(pos + 1, now * 10 + i);
}
inline void solve(int pos, long long now) {
  if (now > n) return;
  if (pos > 1) {
    if (10ll * now * MX > n && 1ll * (now + 1) * MX - 1 <= n) {
      for (int i = 0; i < 7; ++i) {
        int x = ((c - 1ll * now * pw[i]) % mod1 + mod1) % mod1;
        int p =
            lower_bound(ve[i].begin(), ve[i].end(), mod1 - x) - ve[i].begin();
        ans = add(ans, add(1ll * x * ve[i].size() % mod2, sum[i], mod2), mod2);
        ans = dec(ans, 1ll * mod1 * (ve[i].size() - p) % mod2, mod2);
      }
      c += tot;
      return;
    }
    ++c, ans = add(ans, ((c - now) % mod1 + mod1) % mod1, mod2);
  }
  for (int i = (pos == 1 ? 1 : 0); i <= 9; ++i) solve(pos + 1, now * 10 + i);
}
int main() {
  scanf("%lld", &n);
  pw[0] = 1;
  for (int i = 1; i < 9; ++i) pw[i] = 10 * pw[i - 1];
  dfs(1, 0);
  for (int i = 0; i < 7; ++i) sort(ve[i].begin(), ve[i].end());
  solve(1, 0);
  printf("%d\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp 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 mod1 = 998244353, mod2 = 1e9 + 7;
const int N = 2e6, MX = 1e6;
long long n, c;
int tot, ans, lim, pw[N], sum[N];
vector<int> ve[N];
inline int dec(int x, int y, int mod) {
  return (x - y < 0) ? x - y + mod : x - y;
}
inline int add(int x, int y, int mod) {
  return (x + y >= mod) ? x + y - mod : x + y;
}
inline void dfs(int pos, int now) {
  if (now <= n) {
    ++tot;
    int x = dec(tot, now, mod1);
    ve[pos - 1].push_back(x);
    sum[pos - 1] = add(sum[pos - 1], x, mod2);
  }
  if (pos > 6) return;
  for (int i = 0; i <= 9; ++i) dfs(pos + 1, now * 10 + i);
}
inline void solve(int pos, long long now) {
  if (now > n) return;
  if (pos > 1) {
    if (10ll * now * MX > n && 1ll * (now + 1) * MX - 1 <= n) {
      for (int i = 0; i < 7; ++i) {
        int x = ((c - 1ll * now * pw[i]) % mod1 + mod1) % mod1;
        int p =
            lower_bound(ve[i].begin(), ve[i].end(), mod1 - x) - ve[i].begin();
        ans = add(ans, add(1ll * x * ve[i].size() % mod2, sum[i], mod2), mod2);
        ans = dec(ans, 1ll * mod1 * (ve[i].size() - p) % mod2, mod2);
      }
      c += tot;
      return;
    }
    ++c, ans = add(ans, ((c - now) % mod1 + mod1) % mod1, mod2);
  }
  for (int i = (pos == 1 ? 1 : 0); i <= 9; ++i) solve(pos + 1, now * 10 + i);
}
int main() {
  scanf("%lld", &n);
  pw[0] = 1;
  for (int i = 1; i < 9; ++i) pw[i] = 10 * pw[i - 1];
  dfs(1, 0);
  for (int i = 0; i < 7; ++i) sort(ve[i].begin(), ve[i].end());
  solve(1, 0);
  printf("%d\n", ans);
  return 0;
}

```