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>
using namespace std;
int p[11],c[11];
int d,g;
int mn = 1000000000;
int dfs(int n,int np,int nw,int ko){
  if(d+1 == n){
    //cout << n << " " << np << " " << nw << " " << ko << endl;
    if(nw >= g){
      mn = min(mn,ko);
    }else{
      if(p[np]*np+nw >= g){
        int tmp = ((g-nw)%np == 0 ? (g-nw)/np :(g-nw)/np+1);
        mn = min(mn,ko+tmp);
      }
    }
    return mn;
  }
  return min(mn,min(dfs(n+1,np,nw+p[n]*n+c[n],ko+p[n]),dfs(n+1,n,nw,ko)));

}
int main(){
  cin>>d>>g;g/=100;
  for(int i = 1; d >= i; i++){
    cin>>p[i]>>c[i];
    c[i] /= 100;
  }
  cout << dfs(1,0,0,0) << endl;//今の状態,解いてないもので一番高い点数,今の点数

}