#include <bits/stdc++.h>
using namespace std;
const long long mod1 = 998244353;
const long long mod2 = 1e9 + 7;
long long ans;
long long n;
long long cnt1;
vector<long long> val[7];
long long sum[7];
void dfs1(long long k1, long long k2) {
  if (k1 == 7) return;
  val[k1].push_back((cnt1 - k2 + mod1) % mod1);
  sum[k1] += (cnt1 - k2 + mod1) % mod1;
  cnt1++;
  for (long long i = 0; i <= 9; ++i) {
    dfs1(k1 + 1, k2 * 10 + i);
  }
}
bool check(long long x) {
  const long long half = 1000000;
  return x * half + 999999 <= n && x * half * 10 > n;
}
long long cnt2;
long long poww[7] = {1, 10, 100, 1000, 10000, 100000, 1000000};
void dfs2(long long k1, long long k2) {
  if (k2 > n) return;
  if (check(k2)) {
    for (long long i = 0; i <= 6; ++i) {
      long long offset = ((cnt2 + 1 - (k2 * poww[i])) % mod1 + mod1) % mod1;
      long long pos = lower_bound(val[i].begin(), val[i].end(), mod1 - offset) -
                      val[i].begin();
      (ans += (sum[i] + offset * val[i].size() -
               mod1 * (val[i].size() - pos))) %= mod2;
    }
    cnt2 += cnt1;
  } else {
    cnt2++;
    (ans += ((cnt2 - k2) % mod1 + mod1) % mod1) %= mod2;
    for (long long i = (!k1); i <= 9; ++i) {
      dfs2(k1 + 1, k2 * 10 + i);
    }
  }
}
signed main() {
  scanf("%lld", &n);
  dfs1(0, 0);
  for (long long i = 0; i <= 6; ++i) sort(val[i].begin(), val[i].end());
  cnt2 = -1;
  dfs2(0, 0);
  printf("%lld\n", ans);
}

### Prompt

Generate 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 long long mod1 = 998244353;
const long long mod2 = 1e9 + 7;
long long ans;
long long n;
long long cnt1;
vector<long long> val[7];
long long sum[7];
void dfs1(long long k1, long long k2) {
  if (k1 == 7) return;
  val[k1].push_back((cnt1 - k2 + mod1) % mod1);
  sum[k1] += (cnt1 - k2 + mod1) % mod1;
  cnt1++;
  for (long long i = 0; i <= 9; ++i) {
    dfs1(k1 + 1, k2 * 10 + i);
  }
}
bool check(long long x) {
  const long long half = 1000000;
  return x * half + 999999 <= n && x * half * 10 > n;
}
long long cnt2;
long long poww[7] = {1, 10, 100, 1000, 10000, 100000, 1000000};
void dfs2(long long k1, long long k2) {
  if (k2 > n) return;
  if (check(k2)) {
    for (long long i = 0; i <= 6; ++i) {
      long long offset = ((cnt2 + 1 - (k2 * poww[i])) % mod1 + mod1) % mod1;
      long long pos = lower_bound(val[i].begin(), val[i].end(), mod1 - offset) -
                      val[i].begin();
      (ans += (sum[i] + offset * val[i].size() -
               mod1 * (val[i].size() - pos))) %= mod2;
    }
    cnt2 += cnt1;
  } else {
    cnt2++;
    (ans += ((cnt2 - k2) % mod1 + mod1) % mod1) %= mod2;
    for (long long i = (!k1); i <= 9; ++i) {
      dfs2(k1 + 1, k2 * 10 + i);
    }
  }
}
signed main() {
  scanf("%lld", &n);
  dfs1(0, 0);
  for (long long i = 0; i <= 6; ++i) sort(val[i].begin(), val[i].end());
  cnt2 = -1;
  dfs2(0, 0);
  printf("%lld\n", ans);
}

```