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 <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[15],b[15],f[15][1000005];
int main(void)
{
	memset(f,0x3F,sizeof(f));f[0][0]=0;
	int i,j,k,n,m,ans=0x3F3F3F3F;
	scanf("%d%d",&n,&m);m/=100;
	for(i=1;i<=n;++i)scanf("%d%d",&a[i],&b[i]),b[i]/=100;
	for(i=1;i<=n;++i)
	{
		for(j=a[i]-1;j>=0;--j)
		{
			for(k=1000000;k>=i*j;--k)f[i][k]=min(f[i][k],f[i-1][k-i*j]+j);
		}
		for(j=1000000;j>=i*a[i]+b[i];--j)f[i][j]=min(f[i][j],f[i-1][j-i*a[i]-b[i]]+a[i]);
	}
	for(j=m;j<=1000000;++j)ans=min(ans,f[n][j]);
	printf("%d\n",ans);
	return 0;
}