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.Arrays;
import java.util.Scanner;

public class Main 
{
	static long best = 9999999999999l;
	static void go(int[] or)
	{
//		System.out.println("Try: " +Arrays.toString(or));
		
		long cur = 0;
		long remtar = g;
		for(int i = 0; i < d && remtar > 0; i++)
		{
			long nobon = remtar/((or[i]+1)*100);
			if(remtar%((or[i]+1)*100)!=0)nobon++;
			
			if(nobon < pro[or[i]])
			{ //do this many problems only
				cur += nobon;
				break;
			}
			else
			{
				cur += pro[or[i]];
				remtar -= pro[or[i]] * ((or[i]+1)*100);
				remtar -= bon[or[i]];
			}
		}
		best = Math.min(cur, best);
	}

	static void permutate(int[] cur, int i)
	{
		com++;
		for(int j = i; j < d; j++)
		{
			int[] cl = cur.clone();
			if(i!=j)
			{ //swap i and j
				cl[i] = cl[i] + cl[j];
				cl[j] = cl[i] - cl[j];
				cl[i] = cl[i] - cl[j];
			}
			permutate(cl, i+1);
		}
		
		if(i==d)
		{
			tot++;
//			System.out.println(i +" after: " +Arrays.toString(cur));
			go(cur);
		}
	}
	
	static int d;
	static int tot = 0;
	static int com = 0;
	static int g;
	static long[] bon;
	static long[] pro;

	public static void main(String[] args) 
	{
		Scanner sc = new Scanner(System.in);
		d = sc.nextInt();
		g = sc.nextInt(); //target score
		bon = new long[d];
		pro = new long[d];
		
		for(int i = 0; i < d; i++)
		{
			pro[i] =  sc.nextInt();
			bon[i] = sc.nextInt();
		}
		int[] init = new int[d];
		for(int i = 0; i < d; i++)
			init[i] = i;
		
		permutate(init, 0);
		
//		System.out.println("Test : " + Arrays.toString(init));
//		System.out.println(tot + " " + com);
		
		System.out.println(best);
	}
}