Problem:
Score : 300 points

Problem Statement

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).

Sample Input 1

2 700
3 500
5 800

Sample Output 1

3

In this case, there are three problems each with 100 points and five problems each with 200 points. The perfect bonus for solving all the 100-point problems is 500 points, and the perfect bonus for solving all the 200-point problems is 800 points. Takahashi's objective is to have a total score of 700 points or more.

One way to achieve this objective is to solve four 200-point problems and earn a base score of 800 points. However, if we solve three 100-point problems, we can earn the perfect bonus of 500 points in addition to the base score of 300 points, for a total score of 800 points, and we can achieve the objective with fewer problems.

Sample Input 2

2 2000
3 500
5 800

Sample Output 2

7

This case is similar to Sample Input 1, but the Takahashi's objective this time is 2000 points or more. In this case, we inevitably need to solve all five 200-point problems, and by solving two 100-point problems additionally we have the total score of 2000 points.

Sample Input 3

2 400
3 500
5 800

Sample Output 3

2

This case is again similar to Sample Input 1, but the Takahashi's objective this time is 400 points or more. In this case, we only need to solve two 200-point problems to achieve the objective.

Sample Input 4

5 25000
20 1000
40 1000
50 1000
30 1000
1 1000

Sample Output 4

66

There is only one 500-point problem, but the perfect bonus can be earned even in such a case.

Solution in Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

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 min =1001; int count =0; int ans =0;
    int a = 4;
    for (int i =0; i<Math.pow(2,D); i++) {
      String bit = String.format("%" + D + "s", Integer.toBinaryString(i)).replace(' ', '0');
      int N = bit.length();
      for (int j=0; j<N; j++) {
        if (bit.charAt(j) == '1' ) {
          count += p[j] * (j + 1) * 100 + c[j];
          ans += p[j];
        }
      }
      if (count<G) {
        for (int j=N-1; j>=0; j--) {
          if (bit.charAt(j) == '0' ) {
            int l = p[j];
            while (count < G && l > 0) {
              count += (j + 1) * 100;
              ans++;
              l--;
            }
          }
        }
      }
      if (count >= G) {
        min = Math.min(ans,min);
      }
      count =0;
      ans =0;
    }
    System.out.println(min);
  }
}