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(){
  long d,g;
   cin >> d >> g;
  
  vector<long> p(d),c(d);
  long res=0;
  for(long i=0;i<d;i++){
    cin >> p[i] >> c[i];
    res+=p[i];
  }

  for(long i=0;i<(1<<d);i++){
    long score=0;
    long num=0;
    long rest=-1;
   for(long j=0;j<d;j++){
     if((i>>j)&1) {
       score+=(p[j]*100*(j+1)+c[j]);
       num+=p[j];
     }
     else{
      rest=j;
     } 
     }
    if(g>score){
      long need=min((g-score-1)/(100*(rest+1))+1,p[rest]);
       score+=100*(rest+1)*need;
       num+=need;
     }
    if(score>=g) res=min(res,num);
  }
  
  
  
cout << res << endl;

}
