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 <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
int main(void){
  int d,g,i,j,p[11],c[11],t,r,ans=1e9;
  cin >> d >> g;
  for (i=0;i<d;i++) cin >> p[i] >> c[i];
  for (i=0;i<(1<<d);i++){
    t=0;
    r=g;
    for (j=0;j<d;j++) if ((i>>j)%2==1){
      r-=(j+1)*100*p[j]+c[j];
      t+=p[j];
    }
    for (j=d-1;j>=0 && r>0;j--){
      if ((i>>j)%2==1) continue;
      if (r>=(j+1)*100*(p[j]-1)){
        r-=(j+1)*100*(p[j]-1);
        t+=p[j]-1;
      }else{
        t+=(r+(j+1)*100-1)/((j+1)*100);
        r=-100;
      }
      if (r<=0) break;
    }
    if (r<=0 && ans>t) ans=t;
  }
  cout << ans << endl;
  return 0;
}