CodeForces - 121A Lucky Sum（dfs）

Petya loves lucky numbers. Everybody knows that lucky numbers are positive integers whose decimal representation contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Let next(x) be the minimum lucky number which is larger than or equals x. Petya is interested what is the value of the expression next(l) + next(l + 1) + ... + next(r - 1) + next(r). Help him solve this problem.

Input

The single line contains two integers l and r (1 ≤ l ≤ r ≤ 109) — the left and right interval limits.

Output

In the single line print the only number — the sum next(l) + next(l + 1) + ... + next(r - 1) + next(r).

Please do not use the %lld specificator to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specificator.

Examples

Input

2 7

Output

33

Input

7 7

Output

7

Note

In the first sample: next(2) + next(3) + next(4) + next(5) + next(6) + next(7) = 4 + 4 + 4 + 7 + 7 + 7 = 33

In the second sample: next(7) = 7

（思路：首先把所有的幸运数全搜出来，然后排序。然后，只需要遍历搜到的幸运数数组，把包含在区间内的幸运数加起来即可。代码里有两种求法。当然，思路很麻烦的是我的。）

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N =3000;
ll en=4444444444;
ll a[N];
int cnt;
void dfs(ll x)
{
	if(x>en) return ;
	a[++cnt]=x;
	dfs(x*10+4);
	dfs(x*10+7);
}
ll solve(ll x)
{
	ll res=0;
	for(int i=1;i<=cnt;i++)
	{
		if(x>a[i])
		{
			res+=(a[i]-a[i-1])*a[i];
		}
		else
		{
			res+=(x-a[i-1])*a[i];
			return res;
		}
	}
}
int main(void)
{
	cnt=0;
	dfs(4);
	dfs(7);
	sort(a+1,a+1+cnt);
	ll l,r,index;
	ll ans=0,pre;
	cin>>l>>r;
	index=lower_bound(a+1,a+1+cnt,l)-a;
	pre=l;
	if(r>a[index])
	{
		ans+=(a[index]-pre+1)*a[index];
		pre=a[index];
	}
	while(r>a[index])
	{
		ans+=(a[index]-pre)*a[index];
		pre=a[index++];
	}
	if(l>a[index-1])
	{
		ans+=(r-l+1)*a[index];
	}
	else
	{
		ans+=(r-a[index-1])*a[index];
	}
	//cout<<solve(r)-solve(l-1)<<endl;
	cout<<ans<<endl;
	return 0;
}