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

class Main{

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);			//文字の入力
		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] t = new int[n];
		int[] d = new int[n];

		ArrayList<Pair> pairs = new ArrayList<Pair>();
		for(int i = 0;i < n;i++){
			t[i] = sc.nextInt();
			d[i] = sc.nextInt();
			Pair pair = new Pair();
			pair.from = t[i];
			pair.end = d[i];
			pairs.add(pair);
		}

		Collections.sort(pairs,Comparator.reverseOrder());

		int[] cc = new int[100001];			

		

		ArrayList<Integer> one = new ArrayList<Integer>();
		ArrayList<Integer> zero = new ArrayList<Integer>();
		
		cc[pairs.get(0).from]++;			//各寿司について出てきているか判定
		pairs.get(0).from = 1;
		one.add(pairs.get(0).end);
		
		for(int i = 1;i < n;i++){
			if(cc[ pairs.get(i).from ] > 0){			//すでに最高値がある時
				pairs.get(i).from = 0;
				zero.add(pairs.get(i).end);
			}else{
				cc[pairs.get(i).from]++;				//新しい寿司の中で最高値
				pairs.get(i).from = 1;
				one.add(pairs.get(i).end);
			}

		}
	int	syuruimax = one.size();

		long ans = 0;
		int syurui = 0;
		PriorityQueue<Integer> queue = new PriorityQueue<Integer>();

		for(int i = 0;i < k;i++){
			ans += pairs.get(i).end;
			if(pairs.get(i).from == 0){			//最大値じゃない時,
				queue.add(pairs.get(i).end);
				zero.remove(0);
			}else{
				syurui++;
				one.remove(0);
			}
		}	
//
//
//		long[] maxans = new long[syuruimax+1];
//		maxans[syurui] = ans;
//
//		while(syurui < syuruimax){
//			ans += one.get(0);
//			one.remove(0);
//			long del =	queue.poll();
//			ans -= del;
//			syurui++;
//			maxans[syurui] = ans;
//		}
//
//		for(int i = 0;i <= syuruimax;i++){
//			ans = Math.max(ans, maxans[i] + i*i);
//		}

		System.out.println(ans);








	}


}

class Pair implements Comparable{
	int from;		//p
	int end;		//y
	int num;
	int bango;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return end - otherpair.end;
	}
}