Problem:
Score : 900 points

Problem Statement

For a positive integer n, let us define f(n) as the number of digits in base 10.

You are given an integer S.
Count the number of the pairs of positive integers (l, r) (l \leq r) such that f(l) + f(l + 1) + ... + f(r) = S, and find the count modulo 10^9 + 7.

Constraints

1 \leq S \leq 10^8

Input

Input is given from Standard Input in the following format:

S

Output

Print the answer.

Sample Input 1

1

Sample Output 1

9

There are nine pairs (l, r) that satisfies the condition: (1, 1), (2, 2), ..., (9, 9).

Sample Input 2

2

Sample Output 2

98

There are 98 pairs (l, r) that satisfies the condition, such as (1, 2) and (33, 33).

Sample Input 3

123

Sample Output 3

460191684

Sample Input 4

36018

Sample Output 4

966522825

Sample Input 5

1000

Sample Output 5

184984484

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//It is made by Awson on 2018.2.3
#include <bits/stdc++.h>
#define LL long long
#define dob complex<double>
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('\n'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int l1 = 10000000;
const int l2 = 25500000;
const int yzh = 1e9+7;
void read(int &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }
void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

int s, f[l2+5];

int quick_pow(int a, int b) {
    int ans = 1;
    while (b) {
	if (b&1) ans = 1ll*ans*a%yzh;
	a = 1ll*a*a%yzh, b >>= 1;
    }
    return ans;
}
int count1(int n) {
    int ans = 0, r = 0, cnt = 0;
    for (int i = 1; i < l1; i++) {
	cnt -= f[i-1];
	while (cnt+f[r+1] <= s && r < l2) cnt += f[++r];
	if (cnt == s) ++ans;
	if (r == l2) break;
    }
    return ans;
}
int count2(int n) {
    int lim = n/8, ans = lim;
    for (int t = 1; t <= lim; t++)
	if (n%t == 0) {
	    int len = n/t;
	    (ans += (1ll*quick_pow(10, len-1)*9%yzh-t)%yzh) %= yzh;
	}
    return ans;
}
void work() {
    for (int i = 1, r = 10, cnt = 1; i < l1; i++, i = r, r = r*10, cnt++)
	for (int j = i; j < r; j++) f[j] = cnt;
    for (int i = l1; i <= l2; i++) f[i] = 8;
    read(s);
    writeln(((count1(s)+count2(s))%yzh+yzh)%yzh);
}
int main() {
    work();
    return 0;
}