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; cin >> d >> g;
  vector<int> p(d), c(d);
  for(int i = 0; i < d; i++){
    cin >> p[i] >> c[i];
  }
  int ans = 1001001001;
  for(int i = 0; i < (1<<d); i++){
    int sum = 0, ins = 0, num = 0;
    for(int j = 0; j < d; j++){
      if(!(i & (1<<j))) ins = j;
      else{
        sum += (j+1)*100*p[j] + c[j];
        num += p[j];
      }
    }
    if(sum < g){
      if(sum + (ins+1)*100*(p[ins]-1) < g) continue;
      else num += ((g-sum)+((ins+1)*100-1))/((ins+1)*100);
    }
    ans = min(ans, num);
  }
  cout << ans << endl;
}