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:
#ifndef ___CLASS_MODINT
#define ___CLASS_MODINT

#include <cstdint>

template<std::uint32_t mod>
class modint {
private:
	std::uint32_t n;
public:
	modint() : n(0) {};
	modint(std::uint64_t n_) : n(n_ % mod) {};
	bool operator==(const modint& m) const { return n == m.n; }
	bool operator!=(const modint& m) const { return n != m.n; }
	std::uint32_t get() const { return n; }
	modint& operator+=(const modint& m) { n += m.n; n = (n < mod ? n : n - mod); return *this; }
	modint& operator-=(const modint& m) { n += mod - m.n; n = (n < mod ? n : n - mod); return *this; }
	modint& operator*=(const modint& m) { n = std::uint64_t(n) * m.n % mod; return *this; }
	modint operator+(const modint& m) const { return modint(*this) += m; }
	modint operator-(const modint& m) const { return modint(*this) -= m; }
	modint operator*(const modint& m) const { return modint(*this) *= m; }
	modint binpow(std::uint64_t b) const {
		modint ans = 1, m = modint(*this);
		while (b) {
			if (b & 1) ans *= m;
			m *= m;
			b >>= 1;
		}
		return ans;
	}
	modint inv() { return (*this).binpow(mod - 2); }
};

#endif // ___CLASS_MODINT

#include <vector>
#include <iostream>
using namespace std;
using modulo = modint<1000000007>;
long long tenpow(int X) {
	return X == 0 ? 1 : tenpow(X - 1) * 10;
}
const int first_lim = 7;
int main() {
	int N;
	cin >> N;
	int primary_lim = tenpow(first_lim);
	int second_lim = primary_lim + N / (first_lim + 1) + 2;
	vector<long long> SEQ(second_lim);
	long long mul = 1;
	for (int i = 1; i <= first_lim + 1; ++i) {
		for (int j = mul; j < mul * 10 && j < second_lim; ++j) {
			SEQ[j] = i;
		}
		mul *= 10;
	}
	modulo ans = 0;
	int r = 1, s = 0;
	for (int i = 1; i < primary_lim; ++i) {
		while (s < N) s += SEQ[r++];
		if (s == N) ans += 1;
		s -= SEQ[i];
	}
	for (int i = 2; i <= N; ++i) {
		if (N % i == 0) continue;
		if (N / i <= first_lim) break;
		ans += 1;
	}
	for (int i = first_lim + 1; i <= N / 2; ++i) {
		if (N % i == 0) {
			ans += modulo(10).binpow(i - 1) * 9 - modulo(N / i) + 1;
		}
	}
	if (N >= first_lim + 1) ans += modulo(10).binpow(N - 1) * 9;
	cout << ans.get() << endl;
	return 0;
}