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();
    int g = sc.nextInt();
    int[] p = new int[d];
    int[] c = new int[d];
    for(int i = 0; i < d; i++) {
      p[i] = sc.nextInt();
      c[i] = sc.nextInt();
    }
    int ans = Integer.MAX_VALUE;
    if((100 * d * (p[d - 1] - 1)) >= g) {
      if((g % (100 * d)) == 0) {
        ans = Math.min(ans, g / (100 * d));
      } else {
        ans = Math.min(ans, (g / (100 * d)) + 1);
      }
    }
    for(int i = 1; i < Math.pow(2, d); i++) {
      int sum = 0;
      int s = 0;
      int q = 0;
      for(int j = 0; j < d; j++) {
        if((i & (1 << j)) != 0) {
          s += p[j];
          sum += ((100 * (j + 1) * p[j]) + c[j]);
        } else {
          q = j;
        }
      }
      if(sum >= g) ans = Math.min(ans, s);
      if((sum + (100 * (q + 1) * (p[q] - 1))) >= g) {
        if(g > sum) {
          int r = g - sum;
          if((r % (100 * (q + 1))) == 0) {
            s += (r / (100 * (q + 1)));
          } else {
            s += (r / (100 * (q + 1)) + 1);
          }
          ans = Math.min(ans, s);
        }
      }
    }
    System.out.println(ans);
  }
}