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<cstdio>
#include<algorithm>
using namespace std;typedef long long ll;
int n;int p[20];int c[20];int g;int res=0x3f3f3f3f;
int main()
{
	scanf("%d%d",&n,&g);g/=100;
	for(int i=0;i<n;i++)scanf("%d%d",&p[i],&c[i]);
	for(int i=0;i<n;i++)c[i]/=100;
	for(int i=0;i<(1<<n);i++)
	{
		int tot=0;int ret=0;
		for(int j=0,t=i;t;t>>=1,j++)tot+=(t&1)*(p[j]*(j+1)+c[j]),ret+=(t&1)*p[j];
		//printf("i=%d,tot=%d\n",i,tot);
		if(tot>=g)res=min(ret,res);
		else 
		{
			for(int j=n-1;j>=0;j--)
				if(((i>>j)&1)==0)
				{
					if(tot+p[j]*(j+1)<g);
					else {ret+=(g-tot+j)/(j+1);res=min(res,ret);}
					goto ed;
				}
		}ed:;
	}printf("%d",res);return 0;
}