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.Arrays;
import java.util.Scanner;

class Main {
	public static void main(String[] args) {
		new Main().run();
	}

	void run() {
		Scanner sc = new Scanner(System.in);
		int D = sc.nextInt();
		long G = sc.nextLong() / 100;
		long[] p = new long[D];
		long[] c = new long[D];
		for (int i = 0; i < D; ++i) {
			p[i] = sc.nextLong();
			c[i] = sc.nextLong() / 100;
		}
		long ans = Long.MAX_VALUE / 3;
		for (int s = 0; s < 1 << D; ++s) {
			long sum1 = 0, sum2 = 0;
			for (int i = 0; i < D; ++i) {
				if ((s & (1 << i)) > 0) {
					sum1 += p[i];
					sum2 += p[i] * (i + 1) + c[i];
				}
			}
			if (sum2 >= G)
				ans = Math.min(ans, sum1);
			for (int j = 0; j < D; ++j) {
				if ((s & (1 << j)) > 0)
					continue;
				long res = p[j];
				long tmp1 = sum1, tmp2 = sum2;
				while (tmp2 < G && res > 0) {
					++tmp1;
					tmp2 += (j + 1);
					--res;
				}
				if (res == 0)
					tmp2 += c[j];
				if (tmp2 >= G)
					ans = Math.min(ans, tmp1);
			}
		}
		System.out.println(ans);
	}

	void tr(Object... objects) {
		System.out.println(Arrays.deepToString(objects));
	}
}