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:
#include <bits/stdc++.h>

#define _overload(_1,_2,_3,name,...) name
#define _rep(i,n) _range(i,0,n)
#define _range(i,a,b) for(int i=int(a);i<int(b);++i)
#define rep(...) _overload(__VA_ARGS__,_range,_rep,)(__VA_ARGS__)

#define _rrep(i,n) _rrange(i,n,0)
#define _rrange(i,a,b) for(int i=int(a)-1;i>=int(b);--i)
#define rrep(...) _overload(__VA_ARGS__,_rrange,_rrep,)(__VA_ARGS__)

#define _all(arg) begin(arg),end(arg)
#define uniq(arg) sort(_all(arg)),(arg).erase(unique(_all(arg)),end(arg))
#define getidx(ary,key) lower_bound(_all(ary),key)-begin(ary)
#define clr(a,b) memset((a),(b),sizeof(a))
#define bit(n) (1LL<<(n))
#define popcount(n) (__builtin_popcountll(n))

using namespace std;

template<class T>bool chmax(T &a, const T &b) { return (a < b) ? (a = b, 1) : 0;}
template<class T>bool chmin(T &a, const T &b) { return (b < a) ? (a = b, 1) : 0;}

using ll = long long;
using R = long double;
const R EPS = 1e-9L; // [-1000,1000]->EPS=1e-8 [-10000,10000]->EPS=1e-7
inline int sgn(const R& r) {return (r > EPS) - (r < -EPS);}
inline R sq(R x) {return sqrt(max(x, 0.0L));}

const int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
const int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};

// Problem Specific Parameter:


ll ans = 0LL;
const ll mod = 1000000007LL;

ll power(ll a, ll n) {
	ll b = 1LL;
	while (n) {
		if (n & 1) b = b * a % mod;
		a = a * a % mod;
		n >>= 1;
	}
	return b;
}

void cnt(int n, int num) {
	const int d = n / num;
	const ll all = 9LL * power(10LL, d - 1) % mod;
	ans = (ans + all  - num) % mod;
}

const int limit = 23000000;
ll f[limit + 1];

int main(void) {
	int n;
	cin >> n;

	ans = n / 8;
	// num
	for (int i = 1; i * i <= n; ++i) {
		if (n % i) continue;
		if (i <= n / 8) cnt(n, i);
		if (i * i == n) continue;
		if (n / i <= n / 8 ) cnt(n, n / i);
	}

	//cout << ans << endl;

	ll add = 0LL, thr = 1LL;
	rep(i, 1, limit + 1) {
		if (i == thr) add++, thr *= 10LL;
		f[i] = add;
	}

	int lft = 1, rht = 1, cur = 0;
	while (f[lft] <= 7) {
		while (rht < limit and cur < n) cur += f[rht++];
		if (cur == n) ans++;
		cur -= f[lft++];
	}

	ans %= mod;
	cout << ans << endl;

	return 0;
}