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: Runtime Error)

import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		// 値を取得
		int d = sc.nextInt();
		int g = sc.nextInt();
		int[] p = new int[d];
		int[] c = new int[d];
		// 組み合わせの2進数用
		int bin = 0;
		// 最終的に出力する結果
		int ans = Integer.MAX_VALUE;

		for (int i = 0; i < d; i++) {
			p[i] = sc.nextInt();
			c[i] = sc.nextInt();
		}

		// 全てのパターンを網羅するために
		bin = (int) Math.pow(2, d);
		for (int i = 0; i < bin; i++) {
			// 計算中の点数
			int point = 0;
			// 計算中の問題数
			int cnt = 0;
			// 完全に解かない問題の中で最も採点の高いものの点数
			int max = 0;

			for (int j = d - 1; j > 0 ; j--) {
				if (Integer.toBinaryString(i).substring(j).startsWith("1")) {
					point += p[j] * (j + 1) * 100 + c[j];
					cnt += p[j];
				} else if (max < j) {
					max = j;
				}
			}
			if (g <= point) {
				if (cnt < ans) {
					ans = cnt;
				}
			} else {
				int x = p[max];
				while (0 < x) {
					point += (max + 1) * 100;
					cnt++;
					x--;
					if (g <= point) {
						if (cnt < ans) {
							ans = cnt;
							break;
						}
					}
				}
			}
		}

		System.out.println(ans);

		sc.close();
	}
}