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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		String[] conf = scanner.nextLine().split(" ");
		int numPoints = Integer.parseInt(conf[0]);
		long goal = Long.parseLong(conf[1]);
		int[] numProblems = new int[numPoints + 1];
		long[] achievablePoints = new long[numPoints + 1];
		boolean[] hasUsed = new boolean[numPoints + 1];

		for(int pIdx = 1; pIdx <= numPoints; pIdx++) {
			String[] inputs = scanner.nextLine().split(" ");
			numProblems[pIdx] = Integer.parseInt(inputs[0]);
			achievablePoints[pIdx] = numProblems[pIdx] * pIdx * 100L + Integer.parseInt(inputs[1]);
			hasUsed[pIdx] = false;
		}

		long totalPoint = 0L;
		int minNumProblem = 0;
		while(totalPoint < goal) {
			double maxPPP = 0d;
			int pIdxOfMaxPPP = 0;
			for(int pIdx = 1; pIdx <= numPoints; pIdx++) {
				double tmpPPP = hasUsed[pIdx] ? 0d : achievablePoints[pIdx] / (double)numProblems[pIdx];
				if(tmpPPP >= maxPPP) {
					pIdxOfMaxPPP = pIdx;
					maxPPP = tmpPPP;
				}
			}
			if(totalPoint + achievablePoints[pIdxOfMaxPPP] <= goal) {
				totalPoint += achievablePoints[pIdxOfMaxPPP];
				minNumProblem += numProblems[pIdxOfMaxPPP];
				hasUsed[pIdxOfMaxPPP] = true;
			} else break;
		}

		long remainder = goal - totalPoint;
		if(remainder > 0L) {
			int tmpMinNumProblem = Integer.MAX_VALUE;
			for(int pIdx = 1; pIdx <= numPoints; pIdx++) {
				int tmpNumProblem;
				if(!hasUsed[pIdx] && achievablePoints[pIdx] >= remainder) {
					if(numProblems[pIdx] * pIdx * 100 >= remainder) tmpNumProblem = (int) ((remainder - 1L) / (pIdx * 100)) + 1;
					else tmpNumProblem = numProblems[pIdx];
					if(tmpNumProblem < tmpMinNumProblem) tmpMinNumProblem = tmpNumProblem;
				}
			}
			minNumProblem += tmpMinNumProblem;
		}

		System.out.println(minNumProblem);
	}
}