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>
int e[11][1001];
int main(){
 int i,j,k,l,m,n,d,g,p[11],c[11],s=0;
 scanf("%d%d",&d,&g);
 g/=100;
 for(i=1;i<=d;i++){
 scanf("%d%d",&p[i],&c[i]);
 c[i]/=100;
}
 for(i=1;i<=d;i++){
  s+=p[i];
  for(j=1;j<=s;j++){
    e[i][j]=e[i-1][j];
    for(k=1;k<p[i];k++){
    if(j>=k){
    if(e[i-1][j-k]+k*i>e[i][j])e[i][j]=e[i-1][j-k]+k*i;
	}
	}
	k=p[i]; 
	if(j>=k){
    if(e[i-1][j-k]+k*i+c[i]>e[i][j])e[i][j]=e[i-1][j-k]+k*i+c[i];
	}
  }
}
for(j=1;j<=s;j++){
for(i=1;i<=d;i++){
	if(e[i][j]>=g){
		printf("%d",j);
		return 0;
	}
}
}
return 0;
}