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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	@SuppressWarnings("resource")
	public static void main(String args[]) {
		Scanner scanner = new Scanner(System.in);
		int d = scanner.nextInt();
		int g = scanner.nextInt();
		int p[] = new int[d];
		int c[] = new int[d];

		for (int i = 0; i < d; i++) {
			p[i] = scanner.nextInt();
			c[i] = scanner.nextInt() + (i + 1) * 100 * p[i];
		}

		int min = 1000000000;
		for (int i = 0; i < Math.pow(2, d); i++) {
			int scr = 0;
			int cnt = 0;
			List<Integer> list = new ArrayList<>();
			for (int j = 0; j < d; j++) {
				if ((i >> j & 1) == 1) {
					scr += c[j];
					cnt += p[j];
					list.add(j);
				}
			}
			if (scr >= g && cnt < min) {
				min = cnt;
				continue;
			}
			for (int j = d - 1; j >= 0; j--) {
				if (!list.contains(j)) {
					for (int k = 0; k < p[j]; k++) {
						scr += (j + 1) * 100;
						cnt++;
						if (scr >= g && cnt < min)
							min = cnt;
					}
				}
			}

		}

		System.out.println(min);

	}
}