#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f == -1 ? ~x + 1 : x;
}
long long n;
vector<long long> ve[11];
long long sum[11];
const long long N = 1e6, mod1 = 998244353, mod2 = 1e9 + 7;
long long cnt, tmp, ans;
long long m1(long long x) { return (x % mod1 + mod1) % mod1; }
void dfs1(long long x, long long v) {
  if (v > n) return;
  if (x > 6) return;
  ++cnt;
  ve[x].push_back(m1(cnt - v));
  sum[x] += m1(cnt - v);
  for (long long i = 0; i < 10; ++i) dfs1(x + 1, v * 10 + i);
}
void dfs2(long long x, long long v) {
  if (v > n) return;
  if (x >= 1) {
    if (v * N * 10 > n && v * N + N - 1 <= n) {
      for (long long i = 0, t = 1; i <= 6; ++i, t *= 10) {
        long long val = m1(cnt - v * t);
        long long pos =
            lower_bound(ve[i].begin(), ve[i].end(), mod1 - val) - ve[i].begin();
        (ans += val * ve[i].size() + sum[i] - mod1 * (ve[i].size() - pos)) %=
            mod2;
      }
      cnt += tmp;
      return;
    }
    ++cnt;
    (ans += m1(cnt - v)) %= mod2;
  }
  for (long long i = (x == 0 ? 1 : 0); i <= 9; ++i) dfs2(x + 1, v * 10 + i);
}
signed main() {
  n = read();
  dfs1(0, 0);
  tmp = cnt;
  cnt = 0;
  for (long long i = 0; i <= 6; ++i) sort(ve[i].begin(), ve[i].end());
  dfs2(0, 0);
  printf("%d\n", ans);
}

### 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;
inline long long read() {
  long long x = 0, f = 1;
  char c = getchar();
  while (!isdigit(c)) {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (isdigit(c)) {
    x = (x << 3) + (x << 1) + (c ^ 48);
    c = getchar();
  }
  return f == -1 ? ~x + 1 : x;
}
long long n;
vector<long long> ve[11];
long long sum[11];
const long long N = 1e6, mod1 = 998244353, mod2 = 1e9 + 7;
long long cnt, tmp, ans;
long long m1(long long x) { return (x % mod1 + mod1) % mod1; }
void dfs1(long long x, long long v) {
  if (v > n) return;
  if (x > 6) return;
  ++cnt;
  ve[x].push_back(m1(cnt - v));
  sum[x] += m1(cnt - v);
  for (long long i = 0; i < 10; ++i) dfs1(x + 1, v * 10 + i);
}
void dfs2(long long x, long long v) {
  if (v > n) return;
  if (x >= 1) {
    if (v * N * 10 > n && v * N + N - 1 <= n) {
      for (long long i = 0, t = 1; i <= 6; ++i, t *= 10) {
        long long val = m1(cnt - v * t);
        long long pos =
            lower_bound(ve[i].begin(), ve[i].end(), mod1 - val) - ve[i].begin();
        (ans += val * ve[i].size() + sum[i] - mod1 * (ve[i].size() - pos)) %=
            mod2;
      }
      cnt += tmp;
      return;
    }
    ++cnt;
    (ans += m1(cnt - v)) %= mod2;
  }
  for (long long i = (x == 0 ? 1 : 0); i <= 9; ++i) dfs2(x + 1, v * 10 + i);
}
signed main() {
  n = read();
  dfs1(0, 0);
  tmp = cnt;
  cnt = 0;
  for (long long i = 0; i <= 6; ++i) sort(ve[i].begin(), ve[i].end());
  dfs2(0, 0);
  printf("%d\n", ans);
}

```