Problem: 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.

Examples

Input

1


Output

9


Input

2


Output

98


Input

123


Output

460191684


Input

36018


Output

966522825


Input

1000


Output

184984484

Solution (in CPP):

Solution:
    //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;
    }