#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<ll> vll;
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define INF 1000000000
#define LLINF 1000000000000000ll
#define sz(x) ((int)(x).size())
#define fi first
#define sec second
#define all(x) (x).begin(),(x).end()
#define sq(x) ((x)*(x))
#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define repn(i,a,n) for(int (i)=(a);(i)<(int)(n);(i)++)
#define EQ(a,b) (abs((a)-(b))<eps)
#define MOD 1000000007ll
template<class T> void chmin(T& a,const T& b){if(a>b)a=b;}
template<class T> void chmax(T& a,const T& b){if(a<b)a=b;}
int f(ll x){
	int res = 0;
	while(x){
		x/=10ll;
		res++;
	}
	return res;
}
ll num(int x){
	x--;
	ll res = 9ll;
	for(int i=0;i<x;i++){
		res=(res*10ll)%MOD;
	}
	return res;
}
int S;
int d[3000000];
int main(){
	cin >> S;
	ll ans = 0ll;
	for(int i=1;i*i<=S;i++){
		if(S%i!=0)continue;
		ll add = num(i)-S/i+1ll;
		if(i>=9||add>0){
			ans += add;
			ans %= MOD;
		}
		if(i==S/i)continue;
		add = num(S/i)-i+1ll,0ll;
		if(i>=9||add>0){
			ans += add;
			ans %= MOD;
		}
	}
	int sum = 0;
	ll r = 1ll;
	for(int i=1;i<1000000;i++){
		while(sum<S){
			sum += f(r++);
		}
		if(sum==S&&f(i)<f(r-1))ans++;
		sum -= f(i);
	}
	ans %= MOD;
	int L = 9000000;
	for(int i=1;i<=S/7;i++){
		if(S%i==0)continue;
		int a = i-(S%i);
		if(S/i==7&&a>L)continue;
		ans++;
	}
	ans %= MOD;
	cout << ans << endl;
	return 0;
}
### Prompt

Please create a solution in CPP to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef vector<int> vi;
typedef vector<ll> vll;
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define INF 1000000000
#define LLINF 1000000000000000ll
#define sz(x) ((int)(x).size())
#define fi first
#define sec second
#define all(x) (x).begin(),(x).end()
#define sq(x) ((x)*(x))
#define rep(i,n) for(int (i)=0;(i)<(int)(n);(i)++)
#define repn(i,a,n) for(int (i)=(a);(i)<(int)(n);(i)++)
#define EQ(a,b) (abs((a)-(b))<eps)
#define MOD 1000000007ll
template<class T> void chmin(T& a,const T& b){if(a>b)a=b;}
template<class T> void chmax(T& a,const T& b){if(a<b)a=b;}
int f(ll x){
	int res = 0;
	while(x){
		x/=10ll;
		res++;
	}
	return res;
}
ll num(int x){
	x--;
	ll res = 9ll;
	for(int i=0;i<x;i++){
		res=(res*10ll)%MOD;
	}
	return res;
}
int S;
int d[3000000];
int main(){
	cin >> S;
	ll ans = 0ll;
	for(int i=1;i*i<=S;i++){
		if(S%i!=0)continue;
		ll add = num(i)-S/i+1ll;
		if(i>=9||add>0){
			ans += add;
			ans %= MOD;
		}
		if(i==S/i)continue;
		add = num(S/i)-i+1ll,0ll;
		if(i>=9||add>0){
			ans += add;
			ans %= MOD;
		}
	}
	int sum = 0;
	ll r = 1ll;
	for(int i=1;i<1000000;i++){
		while(sum<S){
			sum += f(r++);
		}
		if(sum==S&&f(i)<f(r-1))ans++;
		sum -= f(i);
	}
	ans %= MOD;
	int L = 9000000;
	for(int i=1;i<=S/7;i++){
		if(S%i==0)continue;
		int a = i-(S%i);
		if(S/i==7&&a>L)continue;
		ans++;
	}
	ans %= MOD;
	cout << ans << endl;
	return 0;
}
```