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 main(){
  int D, G, p[11], c[11];
  cin >> D >> G;
  for(int i=0; i<D; ++i){
    cin >> p[i] >> c[i]; 
  }
  
  int ans= 1e9;
  for(int mask=0; mask< (1<<D); ++mask){
    int s=0, num=0, rest_max=-1;
    for(int i=0; i<D; ++i){
      if(mask>>i &1){//1ってのが0001みたいなこと。&はアンド演算子。maskと1とを比較して、同じなら1．違うなら0を返す。
        s+= p[i]*100*(i+1)+c[i]; 
        num+= p[i];
      }
      else{
        rest_max= i;
      }
    }
    if(s<G){
      int s1= 100*(rest_max+1);
      int need= (G-s+s1-1)/s1;
      if(need>= p[rest_max]) continue;
      num+= need;     
    }
    ans=min(ans,num);
  }
  cout << ans << endl;
}