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;

int main() {
  int D,G,i,ans=10000;
  cin>>D>>G;
  vector<int> p(D),c(D);
  for(i=0;i<D;i++)cin>>p[i]>>c[i];
  
  for(int tmp=0;tmp<(1<<D);tmp++){
    bitset<10> x(tmp);
    int scr=0,sum=0,m=-1;
    
    for(i=0;i<D;i++){
      if(x.test(i)){
        scr+=100*(i+1)*p[i]+c[i];
        sum+=p[i];
      }else{m=i;}
      if(sum>=ans||scr>=G)break;
    }
    
    if(scr<G&&sum<ans){
      if(scr+p[m]*100*(m+1)>=G){
        sum+=(G-scr+100*m)/(100*(m+1));
        scr=G;
      }
    }
    
    if(scr>=G)ans=min(ans,sum);
    //cout<<sum<<","<<ans<<","<<scr<<endl;
  }
  
  cout<<ans<<endl;
}
