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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Arrays;
import java.util.Comparator;
import java.util.Scanner;

public class Main implements Runnable {

    private static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        new Thread(null, new Main(), "", 128 * 1024 * 1024).start();
    }

    public void run() {
        int n = sc.nextInt() /* <= 10 */, goal = sc.nextInt();
        int[][] arr = new int[n][5];
        for (int i = 0; i < n; i++) {
            arr[i][0] = (i + 1) * 100; // for which score problem
            arr[i][1] = sc.nextInt(); // how many
            arr[i][2] = sc.nextInt(); // bonus
            arr[i][3] = arr[i][0] * arr[i][1] + arr[i][2]; // total
            arr[i][4] = arr[i][3] / arr[i][1]; // efficiency
        }

        // sort by efficiency
        Arrays.sort(arr, Comparator.comparingInt(a -> -a[4]));

        int count = 0;
        int progress = 0;
        int i = 0;
        for (; i < n; i++) {
            if (goal >= arr[i][3] + progress) {
                progress += arr[i][3];
                count += arr[i][1];
            } else {
                break;
            }
        }

        int[][] left = Arrays.copyOfRange(arr, i, arr.length);
        Arrays.sort(arr, Comparator.comparingInt(a -> -a[0]));

        for (int j = 0; j < left.length; j++) {
            for (int k = 0; k < left[j][1]; k++) {
                if (progress < goal) {
                    progress += left[j][0];
                    count++;
                } else {
                    System.out.println(count);
                    return;
                }
            }
            progress += left[j][2];
        }
    }
}