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<cstring>
#define fo(i,a,b)for(int i=a;i<=b;i++)
#define min(a,b)(a<b?a:b)
using namespace std;
int n,g,p[105],c[105],f[2][105505],o,zs,ans;
int main(){
	scanf("%d%d",&n,&g);g/=100;
	fo(i,1,n)scanf("%d%d",&p[i],&c[i]),c[i]/=100;
	fo(i,0,105500)f[0][i]=f[1][i]=1e9;f[0][0]=0;
	fo(i,1,n){
		o^=1;
		fo(j,0,zs)f[o][j]=f[!o][j];
		fo(j,1,p[i]-1)
			fo(k,0,zs)f[o][k+i*j]=min(f[o][k+i*j],f[!o][k]+j);
		fo(k,0,zs)f[o][k+i*p[i]+c[i]]=min(f[o][k+i*p[i]+c[i]],f[!o][k]+p[i]);
		zs+=p[i]*i+c[i];
	}
	ans=1e9;
	fo(i,g,zs)ans=min(ans,f[o][i]);
	printf("%d",ans);
}