#include<bits/stdc++.h>
#define int ll
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
typedef long long ll;
typedef pair<int,int> pii;
#define Rep(i,a,b) for(register int i=(a);i<=int(b);++i)
#define Dep(i,a,b) for(register int i=(a);i>=int(b);--i)
#define rep(i,a,b) for(register int i=(a);i<int(b);++i)
#define mem(x,v) memset(x,v,sizeof(x))
/*inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}*/
#define gc getchar
#define pc putchar
#define fi first
#define se second
#define debug(x) cout << #x" = " << x << endl;
#define pp(x,y) cout << "pp: " << x << " " << y << endl;
#define rank __RANK
inline ll read(){
	register ll x=0,f=1;register char c=gc();
	for(;!isdigit(c);c=gc())if(c=='-')f=-1;
	for(;isdigit(c);c=gc())x=(x<<1)+(x<<3)+(c^48);
	return x*f;
}
#define rd read
void write(ll x){if(x<0)x=-x,pc('-');if(x>=10)write(x/10);putchar(x%10+'0');}
void writeln(ll x){write(x);puts("");}
const int mod = 1e9+7; 
inline int qpow(int a,int b){
	int ans = 1;
	for(;b;b>>=1,a=1ll*a*a%mod)
		if(b&1)ans=1ll*ans*a%mod;
	return ans;
}
#define MN 40000005
int S;
unsigned a[MN];
inline void add(ll &x,int y){(x+=y)>=mod?x-=mod:0;}
ll ans;
signed main(){
	S = rd();
	for(int i=1;i<=4e7;++i) a[i] = a[i/10] + 1;
	int r = 1,sum = 0;
	for(int i=1;i<1e7;++i){
		while(sum < S) sum += a[r],++r;
		if(sum == S) ans = (ans+1)%mod;
		sum -= a[i];
	}
	for(int i=1;i<=S/8;++i){
		if(S%i!=0){
			ans = (ans+1)%mod;
		} else
		if(S%i==0){
			ans = (ans + (qpow(10,S/i) - qpow(10,S/i-1) - i + 1) % mod + mod) % mod;
		}
	}
	writeln(ans);
}
### Prompt

Your challenge is to write a CPP solution 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>
#define int ll
using namespace std;
#pragma comment(linker, "/stack:200000000")
#pragma GCC optimize("Ofast")
#pragma target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
typedef long long ll;
typedef pair<int,int> pii;
#define Rep(i,a,b) for(register int i=(a);i<=int(b);++i)
#define Dep(i,a,b) for(register int i=(a);i>=int(b);--i)
#define rep(i,a,b) for(register int i=(a);i<int(b);++i)
#define mem(x,v) memset(x,v,sizeof(x))
/*inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}*/
#define gc getchar
#define pc putchar
#define fi first
#define se second
#define debug(x) cout << #x" = " << x << endl;
#define pp(x,y) cout << "pp: " << x << " " << y << endl;
#define rank __RANK
inline ll read(){
	register ll x=0,f=1;register char c=gc();
	for(;!isdigit(c);c=gc())if(c=='-')f=-1;
	for(;isdigit(c);c=gc())x=(x<<1)+(x<<3)+(c^48);
	return x*f;
}
#define rd read
void write(ll x){if(x<0)x=-x,pc('-');if(x>=10)write(x/10);putchar(x%10+'0');}
void writeln(ll x){write(x);puts("");}
const int mod = 1e9+7; 
inline int qpow(int a,int b){
	int ans = 1;
	for(;b;b>>=1,a=1ll*a*a%mod)
		if(b&1)ans=1ll*ans*a%mod;
	return ans;
}
#define MN 40000005
int S;
unsigned a[MN];
inline void add(ll &x,int y){(x+=y)>=mod?x-=mod:0;}
ll ans;
signed main(){
	S = rd();
	for(int i=1;i<=4e7;++i) a[i] = a[i/10] + 1;
	int r = 1,sum = 0;
	for(int i=1;i<1e7;++i){
		while(sum < S) sum += a[r],++r;
		if(sum == S) ans = (ans+1)%mod;
		sum -= a[i];
	}
	for(int i=1;i<=S/8;++i){
		if(S%i!=0){
			ans = (ans+1)%mod;
		} else
		if(S%i==0){
			ans = (ans + (qpow(10,S/i) - qpow(10,S/i-1) - i + 1) % mod + mod) % mod;
		}
	}
	writeln(ans);
}
```