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: Accepted)

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        try (Scanner sc = new Scanner(System.in);) {
            new Main().solve(sc);
        }
    }

    void solve(Scanner sc) {
        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;
        for (int i = 0; i < (1 << d); i++) {
            int point = 0;
            int count = 0;
            for (int j = 0; j < d; j++) {
                if (((i >> j) & 1) == 1) {
                    point += c[j] + (j + 1) * 100 * p[j];
                    count += p[j];
                }
            }

            if (point < g) {
                count = plus(p, c, i, d, count, point, g);
            }
            ans = Math.min(ans, count);
        }
        System.out.println(ans);
    }

    int plus(int[] p, int[] c, int i, int d, int count, int point, int g) {
        for (int j = d - 1; j >= 0; j--) {
            if (((i >> j) & 1) == 0) {
                for (int k = 0; k < p[j]; k++) {
                    point += (j + 1) * 100;
                    count++;

                    if (point >= g) {
                        return count;
                    }
                }
            }
        }
        return count;
    }
}