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.Scanner;
public class Main {

	public static void main(String[] args) {
		Scanner in=new Scanner(System.in);
		int n = in.nextInt(), goal = in.nextInt();
		Problem[] p = new Problem[n];
		for (int i=0; i<n; i++) {
			p[i] = new Problem(in.nextLong(), (i+1)*100, in.nextLong());
		}

		//ビット全探索+@
		long min = Long.MAX_VALUE;
		for (int i=0; i<(1<<n); i++) {

			int num = 0; //その組み合わせでは何問解いたか？
			long sum = 0L; //その組み合わせで得られたスコアの合計

			//全完していない問題のうち、最もスコアが高いもの、すなわち
			//i をビット表現したときに最も右側にある 0 の位置を記録する変数
			int rightMost = -1;

			for (int j=n-1; j>=0; j--) { //
				if ((1&i>>j) == 1) {
					sum += p[j].num * p[j].score + p[j].bonus;
					num += p[j].num;
				}
				else if ((1&i>>j) == 0 && rightMost == -1) {
					rightMost = j;
				}
			}

			//sum が目標スコア以上の場合は、解いた問題の数をそのまま min に記録する
			if (sum >= goal) {
				min = Math.min(min, num);
				continue;
			}
			else if (rightMost != -1){
				//そうでない場合かつ、全完していない問題が少なくとも1つある場合

				//目標スコアに届くまで、スコアが最も多い問題を needNum 回解く必要がある
				long needNum = (goal-sum+p[rightMost].score-1) / p[rightMost].score;

				if (needNum > p[rightMost].num) {
					//スコアが最も多い問題が needNum 問より少ないなら、放棄
					continue;
				}
				else {
					//そうでないなら、スコアが最も多い問題を needNum 問解き、
					//min を更新する
					min = Math.min(min, num + needNum);
				}
			}

		}

		System.out.println(min);

	}

	static class Problem {
		long num, score, bonus;
		Problem (long a, long b, long c) {
			num = a; score = b; bonus = c;
		}
	}


}