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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Arrays;
import java.util.Scanner;
import java.util.stream.IntStream;

public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int d = scanner.nextInt(), g = scanner.nextInt();
			Score[] s = new Score[d];
			IntStream.range(0, d).forEach(i -> s[i] = new Score((i + 1) * 100, scanner.nextInt(), scanner.nextInt()));
			Arrays.sort(s);
			System.out.println(IntStream.range(1, 1 << d).map(i -> getProblemNo(i, s, g)).min().getAsInt());
		}
	}

	/**
	 * 目標スコアが達成させるための解く問題数を計算する
	 * 
	 * @param mask 使用する問題の2進数表現
	 * @param s    問題数とスコア数の配列
	 * @param g    目標スコア
	 * @return 目標スコアが達成させるための解く問題数
	 */
	private static int getProblemNo(int mask, Score[] s, int g) {
		int result = 0;
		for (int i = 0; i < s.length; i++) {
			if ((mask & (1 << i)) > 0) {
				if (g > s[i].max) {
					result += s[i].count;
					g -= s[i].max;
				} else {
					int index = Arrays.binarySearch(s[i].scores, g);
					if (index < 0) {
						index = ~index;
					}
					result += index + 1;
					return result;
				}
			}
		}
		// 達成できない場合
		return Integer.MAX_VALUE;
	}

	/**
	 * 問題数とスコアを表すクラス
	 */
	private static class Score implements Comparable<Score> {
		/** 問題数 */
		int count;
		/** スコアの配列 */
		int[] scores;
		/** 完成したときのスコア数 */
		int max;

		/**
		 * コンストラクタ
		 * 
		 * @param n 1問あたりのスコア
		 * @param p 問題数
		 * @param c ボーナススコア
		 */
		Score(int n, int p, int c) {
			scores = new int[count = p];
			IntStream.range(0, p - 1).forEach(i -> scores[i] = (i + 1) * n);
			scores[p - 1] = max = p * n + c;
		}

		@Override
		public int compareTo(Score score) {
			// 問題を完了させたときのスコア数/問題数 で降順でソート
			return Double.compare(((double) score.max) / score.count, ((double) this.max) / this.count);
		}
	}
}