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: Wrong Answer)

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

public class Main {
	private static class ProbSet {
		int eachScore;
		int num;
		int bonus;
		int total;

		public ProbSet(int eachScore, int num, int bonus) {
			this.eachScore = eachScore;
			this.num = num;
			this.bonus = bonus;
			this.total = eachScore * num + bonus;
		}
	}

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String[] tokens = in.nextLine().split(" ");
		int D = Integer.parseInt(tokens[0]);
		int G = Integer.parseInt(tokens[1]);
		int[] p = new int[D];
		int[] c = new int[D];
		for (int i = 0; i < D; ++i) {
			tokens = in.nextLine().split(" ");
			p[i] = Integer.parseInt(tokens[0]);
			c[i] = Integer.parseInt(tokens[1]);
		}
		Comparator<ProbSet> aveBaseComp = new Comparator<ProbSet>() {
			@Override
			public int compare(ProbSet o1, ProbSet o2) {
				double o1Ave = (double)o1.total / (double)o1.num;
				double o2Ave = (double)o2.total / (double)o2.num;
				if (o1Ave < o2Ave) {
					return 1;
				} else if (o1Ave > o2Ave) {
					return -1;
				} else {
					return 0;
				}
			}
		};
		Comparator<ProbSet> eachBaseComp = new Comparator<ProbSet>() {
			@Override
			public int compare(ProbSet o1, ProbSet o2) {
				if (o1.eachScore < o2.eachScore) {
					return 1;
				} else if (o1.eachScore > o2.eachScore) {
					return -1;
				} else {
					return 0;
				}
			}
		};

		List<ProbSet> probSets = new ArrayList<>();
		for (int i = 0; i < D; ++i) {
			int score = (i + 1) * 100;
			probSets.add(new ProbSet(score, p[i], c[i]));
		}
		Collections.sort(probSets, aveBaseComp);
		int solvedNum = 0;
		while (true) {
			ProbSet probSet = probSets.get(0);
			System.out.println("G= " + G);
			if (G > probSet.total) {
				G -= probSet.total;
				solvedNum += probSet.num;
				probSets.remove(0);
			} else {
				int minNum = Integer.MAX_VALUE;
				for (int i = 0; i < probSets.size(); ++i) {
					ProbSet set= probSets.get(i);
					// int n = Math.min(set.num, Math.floorDiv(G, set.eachScore));
					int n = Math.floorDiv(G,  set.eachScore);
					if (minNum > n) {
						minNum = n;
					}
				}
				solvedNum += minNum;
				break;
			}
		}
		System.out.println(solvedNum);
		in.close();
	}
}