A programming competition site AtCode provides algorithmic problems. Each problem is allocated a score based on its difficulty. Currently, for each integer i between 1 and D (inclusive), there are p_i problems with a score of 100i points. These p_1 + … + p_D problems are all of the problems available on AtCode.

A user of AtCode has a value called total score. The total score of a user is the sum of the following two elements:

* Base score: the sum of the scores of all problems solved by the user.
* Perfect bonuses: when a user solves all problems with a score of 100i points, he/she earns the perfect bonus of c_i points, aside from the base score (1 ≤ i ≤ D).



Takahashi, who is the new user of AtCode, has not solved any problem. His objective is to have a total score of G or more points. At least how many problems does he need to solve for this objective?

Constraints

* 1 ≤ D ≤ 10
* 1 ≤ p_i ≤ 100
* 100 ≤ c_i ≤ 10^6
* 100 ≤ G
* All values in input are integers.
* c_i and G are all multiples of 100.
* It is possible to have a total score of G or more points.

Input

Input is given from Standard Input in the following format:


D G
p_1 c_1
:
p_D c_D


Output

Print the minimum number of problems that needs to be solved in order to have a total score of G or more points. Note that this objective is always achievable (see Constraints).

Examples

Input

2 700
3 500
5 800


Output

3


Input

2 2000
3 500
5 800


Output

7


Input

2 400
3 500
5 800


Output

2


Input

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000


Output

66
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
#define rep(i,a,b) for(int i=a;i<=b;i++)
const int N=1e2+5;
LL n,g,p[N],c[N],a[N],ans;
int main()
{
	cin>>n>>g;
	rep(i,1,n)
	{
		cin>>p[i]>>c[i];
		a[i]=i;
	}
	LL ans=1e18;
	do
	{
		LL sum=0,tot=0;
		rep(i,1,n)
		{
			int j=a[i];
			if(sum+p[j]*j*100+c[j]<=g)
			{
				sum+=p[j]*j*100+c[j];
				tot+=p[j];
			}
			else
			{
				LL ad=(g-sum)/(j*100);
				if((g-sum)%(j*100))ad++;
			    tot+=min(ad,p[j]);
				ans=min(ans,tot);
				break;
			}
		}
		ans=min(ans,tot);

	}while(next_permutation(a+1,a+1+n));
	cout<<ans;
	return 0;
}
