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.*;
class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt(); 
    int G = sc.nextInt();
    int[] p = new int[N];
    int[] c = new int[N];
    for (int i = 0; i < N; i++) {
      p[i] = sc.nextInt();
      c[i] = sc.nextInt();
    }
    int min = Integer.MAX_VALUE;
    for (int i = 0; i < Math.pow(2,N); i++) {
      int count = 0;
      int total = 0;
      int t = -1;
      for (int j = N-1; j >= 0; j--) {
        if((1&i>>j) == 1) {
          count += p[j];
          total += c[j] + p[j]*(j+1)*100;
        } else if (t == -1) {
          t = j;
        }
      }
      if (G > total && t != -1) {
        for (int j = 0; j < p[t]-1; j++) {
          total += (t+1)*100;
          count ++;
          if (total >= G) break;
        }
      }
      if (G > total) continue;
      //System.out.println("i: " + i + " total: " + total + " count: " + count + " min: " + min);
      min = Math.min(min, count);
    }
    System.out.println(min);
  }
}