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
import java.util.*;

public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int d = sc.nextInt();//問題番号
    int g = sc.nextInt();//目標点数
    int[] p = new int[d];//各番号の問題数
    int[] c = new int[d];//各番号のボーナス点数
    int min_count = Integer.MAX_VALUE;

    for(int i = 0; i < d; i++){
      p[i] = sc.nextInt();
      c[i] = sc.nextInt();
    }

    for(int i = 0; i < (int)Math.pow(2, d); i++){
      int max_idx = -1;
      int idx = d;
      int cnt = 0;
      int sum = 0;
      for(int j = 0; j < d; j++){
        idx -= 1;
        if((1 & (i >> j)) == 1){
          sum += p[idx] * (idx+1) * 100 + c[idx];
          cnt += p[idx];
        }else{
          max_idx = Math.max(max_idx, idx);
        }
      }
      if(sum < g){
        int one_cnt = 0;
        while( sum < g && one_cnt <= p[max_idx]){
          sum += 100 * (max_idx + 1);
          one_cnt += 1;
        }
        if(sum >= g){
          cnt += one_cnt;
        }else{
          cnt = Integer.MAX_VALUE;
        }
      }
      min_count = Math.min(min_count, cnt);
    }

    System.out.println(min_count);
  }
}
